What is the slope of the line through (3,2) and (-3,4)?

Answer:

D

Step-by-step explanation:

Answer:

a = 2m - b , a = 15

Step-by-step explanation:

(a)

m = [tex]\frac{1}{2}[/tex] (a + b) ← multiply both sides by 2 to clear the fraction

2m = a + b ( subtract b from both sides )

2m - b = a

(b)

when m = 19 and b = 23 , then

a = 2(19) - 23 = 38 - 23 = 15

a. 'a' in terms of m and b is given by: a = 2m - b.
b. when m = 19 and b = 23, the value of 'a' is 15.

a. To express a in terms of m and b, we can rearrange the mean formula:

m = (1/2)(a + b)

Now, we'll solve for 'a':

Multiply both sides by 2 to eliminate the fraction:

2m = a + b

Subtract 'b' from both sides:

2m - b = a

So, 'a' in terms of m and b is given by: a = 2m - b.

b. Now that we have the expression for 'a', we can find its value when m = 19 and b = 23:

a = 2m - b

a = 2(19) - 23

a = 38 - 23

a = 15

So, when m = 19 and b = 23, the value of 'a' is 15.

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Answer:

In order to go from the first point to the second, we need to rise from 4 to 7, i.e. by 3. We also need to run from -4 to -2, i.e. by 2. Therefore the slope is 32 .

Step-by-step explanation:

a) The variables are:

b)

x + y ≥ 12

x*$0.75 + y*$1.25 ≤ $15

a) First we need to define the variables, in this case we have:

b) Now we want to write the system, we know that he must buy at least 12 items, then:

x + y ≥ 12

And he can spend, at most, $15, then:

x*$0.75 + y*$1.25 ≤ $15

These two inequalities conform the system of inequalities:

x + y ≥ 12

x*$0.75 + y*$1.25 ≤ $15

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Answer:

2:7

Step-by-step explanation:

If 2 of the puppies are brindle that leaves 7 that are not out of the nine making the ratio 2:7.

The exponential equation that models this situation is:

[tex]V(t) = 18000(0.85)^t, 0 \leq t \leq 5[/tex].

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

For this problem, the parameters are:

A(0) = 18000, r = 0.15.

Hence the equation for the value of the car over the next 5 years is:

[tex]A(t) = A(0)(1 - r)^t[/tex]

[tex]A(t) = 18000(1 - 0.15)^t[/tex]

[tex]V(t) = 18000(0.85)^t, 0 \leq t \leq 5[/tex].

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Answer:

the only one digit number that is cube and square would be 1

Step-by-step explanation:

because 1 x1 = 1 and 1×1×1= 1

no other one digit number is both a square number and cubic number :)

hope it helps please let me know if you need more help :)

F is located on the number 4.5 on the number line.

We know that point D is at -6

Point E is at 8.

Point F is between D and E, such that the ratio:

DF:FE is 3:1

So if we divide the distance between D and E in 4 parts, 3 of these parts are DF, and one of these parts is FE.

First, the distance between E and D is:

distance = 8 - (-6) = 14 units.

Now, if we divide that by 4, we get:

14/4 = 3.5

Then we have:

DF = 3*(3.5) = 10.5

This means that F is at 10.5 units to the right of D, then:

F = D + 10.5 = -6 + 10.5 = 4.5

F is located on the number 4.5 on the number line.

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Answer:

4.5 is the answer :D

Step-by-step explanation:

The width of the freezer is 2.25ft

The formula for calculating the volume of rectangular prism is expressed as:

V = lwh

Given the following

length l = 8feet

w is the width

height h = 3feet

Substitute

54 = 8*3w

54 = 24w

w = 54/24

w = 2.25ft

Hence the width of the freezer is 2.25ft

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Answer:

37.44 cm²

Step-by-step explanation:

To find the width, follow these steps.

10.4*2=20.8

28-20.8=7.2

7.2/2=3.6

The width is 3.6.

To find the area, multiply the width and the length together.

10.4*3.6=37.44

The area of the rectangle is 37.44 cm²

Hope this helps!

Answer:

150 [tex]m^{2}[/tex]

Step-by-step explanation:

Consider the figure as an isosceles triangle and a rectangle combined.

ISOSCELES TRIANGLE

[tex]A = \frac{1}{2} bh[/tex]

[tex]A = \frac{1}{2} *15*12[/tex]

A = 90

RECTANGLE

[tex]A=lw[/tex]

[tex]A = 5*12[/tex]

[tex]A = 60[/tex]

60 + 90 = 150

Answer:

150m

Step-by-step explanation:

break up the polygon into a triangle and a rectangle.

Triangle:

to find the height of the triangle, subtract 5 from 20

height = 15m     base = 12m

[tex]area=\frac{1}{2}bh\\ A=\frac{1}{2} (12)(15)\\A=6(15)\\A=90m[/tex]

rectangle:

[tex]A=bh\\A=12(5)\\A=60m[/tex]

Add both areas together:

[tex]90+60=150m[/tex]

Answer: Isosceles Trapezoid

Step-by-step explanation:

See the graph of the quadrilateral.

The side LO is congruent to the side MN, the diagonal LN is congruent to the diagonal MO, and the angle L is congruent to the angle M in an isosceles trapezoid, denoted by the symbols LMNO.

What are the conditions for an Isosceles Trapezoid?

The conditions listed below demonstrate that any trapezoid is an isosceles trapezoid:

When these conditions are met by the given trapezoid LMNO, it will be referred to as an isosceles trapezoid. Hence, the following conditions of trapezoid LMNO need to be fulfilled,

LN ≅ MO

LO ≅ MN

∠L ≅ ∠M

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✔ (A.) LN ≅ MO

✔ (C.) LO ≅ MN

✔ (E.) ∠L ≅ ∠M

Answer:

D. continuous

The length of the side QR in the task content can be determined as; 12.22cm.

It follows from the task content that 2 lengths and one included angle are given in the task content. It therefore follows that the length of side QR can be evaluated by the cosine rule as follows;

QR² = 13² + 4² - (2×4×13×cos70)

QR = √149.4

QR = 12.22cm

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x<3

Step-by-step explanation:

–2(5 – 4x) < 6x – 4

(–10 + 8x) < 6x – 4

8x –6x < –4 + 10

2x < 6

x < 3

Tangent of a circle is a line which touches the circle at a point from outside the circle.

Given a tangent to a circle through a point outside the circle.

We have to draw a tangent to a circle from point outside the circle.

Tangent of a circle is a line which touches the circle at a point from outside the circle.. The point at which tangent touches the circle is known as point of contact.The radius of the circle and tangent are perprndicukar to each other at the point of contact. There cannot be any tangent  to the circle through a point inside the circle. There can be one tangent to a point on the circle. The steps of drawing a tangent to a circle are as under:

1) Draw a circle with the radius neede with center O.

2) Join center of the circle O and any point P on the circle OP is the radius of the circle.

3) Draw a line perpendicular to radius OP through point P which is the intersection point of tangent and circle.

This line is the tangent to the circle at P.

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Answer:

igu

you could crop out question n use pic :)

Answer:

[tex]\huge\boxed{\sf 7x + 10\ cm}[/tex]

Step-by-step explanation:

= 5 + 2x + 2x + 5 + 3x

Combine like terms

= 2x + 2x + 3x + 5 + 5

= 7x + 10 cm

[tex]\rule[225]{225}{2}[/tex]

Answer:

7x + 10

Step-by-step explanation:

Perimeter is the sum of all the sides together, the distance around a shape.

3x + 5 + 2x + 2x + 5

Organize so the variables are on one side

3x + 2x + 2x + 5 + 5

Combine like terms:

7x + 10

Your answer would be: 7x + 10

I hope it helps! Have a great day!

bren~

Answer:

f(4) = 5

Step-by-step explanation:

Think about it this way

a point with an x value of 4,

this x value is put into the function,

then the function outputted a y value (or value) of 5.

So, we can name this function f(x)

Then,  we let x = 4,

then the output for f(4) = 5

Answer: f(4) = 5

Step-by-step explanation:

      If the point (4, 5) is on the graph of a function, this means that when you input 4 you get an output of 5.

      Inputs are usually shown with the x-coordinate, and outputs are usually shown with the y-coordinate.

      This means that (4, 5) is an input of 4 for an output of 5.

Answer:

The possible length of the third side is greater that 4ft and less that 20 ft

4<x<20

For finding the given literal equation is necessary to apply the power 2 on both sides of the given equation and after that you should rewrite the equation as a function of the variable m .

There are different power rules, see some them:

1. Multiplication with the same base: you should repeat the base and  add the exponents.

2. Division with the same base: you should repeat the base and  subctract the exponents.

3.Power. For this rule, you should repeat the base and multiply the exponents.

4. Zero Exponent. When you have an exponent equals to zero, the result must be 1.

The question gives: [tex]t=2\pi *\sqrt{\frac{m}{k}}[/tex] and from the previous equation, you should find [tex]m=\frac{kt^2}{4\pi ^2}[/tex].

STEP 1 - Applying the power 2 on both sides of the equation [tex]t=2\pi *\sqrt{\frac{m}{k}}[/tex].

[tex]t=2\pi *\sqrt{\frac{m}{k}}\\ \\ t^2=(2\pi *\sqrt{\frac{m}{k}})^2\\ \\ t^2=4\pi ^2*\frac{m}{k}[/tex]

STEP 2 - Rewriting the equation  [tex]t^2=4\pi ^2*\frac{m}{k}[/tex]  as a function of the variable m .

[tex]t^2=4\pi ^2*\frac{m}{k}\\ \\ t^2k=4\pi ^2m\\ \\ m=\frac{kt^2}{4\pi ^2}[/tex]

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Answer:

1. divide both sides of the equation by 2pi

2. square both sides of the equation

3. use the rule to simplify a fraction to a power

4. multiply both sides of the equation by k

Step-by-step explanation:

edg '22

The smallest number by which 29160 should be divided so that the quotient becomes a perfect cube is 5

The given number is 29160

Note that:

29160 can be factored into 5832

That is, 29160 = 5832 x 5

Take the cube of 5832:

[tex]\sqrt[3]{5832}=18[/tex]

Since it has been confirmed that 5832 is a perfect cube, the smallest number by which 29160 should be divided so that the quotient becomes a perfect cube is 5

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Answer:

b

Step-by-step explanation:

b shows direct variation because we can clearly tell that the slope of the graph is 1/2. The graph is also linear, and it crosses the origin (0,0).

a: This graph can be represented as y=2, and that is not a direct variation.

b: Above explanation. This graph is a direct variation.

c: This graph can be represented as x=-8, and that is not a direct variation.

d: This graph has the slope, but it does not cross the origin, so it is not a direct variation.

Answer:

Graph b

Step-by-step explanation:

Direct variation means "y varies directly as x”:

Direct variation equation:

[tex]y=kx[/tex]

where k is the (non-zero) constant of variation.

Note, therefore, that a direct variation equation passes through the origin,  as when x = 0:  

[tex]\implies y=k(0) \implies y=0[/tex]

Graph a

y does not change as x varies.

Therefore this graph does not model a direct variation.

Graph b

As x gets bigger, so does y.  As x gets smaller, so does y.  

The line passes through the origin (0, 0).

Therefore this graph models a direct variation.

Graph c

x does not change as y varies.

Therefore this graph does not model a direct variation.

Graph d

As x gets bigger, so does y.  As x gets smaller, so does y.  

However, the line does not pass through the origin (0, 0).

Therefore this graph does not model a direct variation.

3) [tex]\angle ADB[/tex] and [tex]\angle BDE[/tex] are supplementary (if two angles form a linear pair, they are supplementary)

4) [tex]\angle ADC[/tex] and [tex]\angle CDE[/tex] are supplementary (if two angles form a linear pair, they are supplementary)

5) [tex]\angle ADB \cong \angle ADC[/tex] (supplements of congruent angles are congruent)

6) [tex]\triangle ADB \cong \triangle ADC[/tex] (SAS)

7) [tex]\angle ABD \cong \angle ACD[/tex] (CPCTC)

Answer:

Domain: All real numbers ; Range: y> -3

Step-by-step explanation:

x = 4 would be the solution of the given problem.

I can't speak to the first part of this question, as I don't totally have context for what they're asking, but we can solve for x using one of the laws of logarithms, namely:

log.m + log.n = log.mn

Using this law, we can combine and rewrite our initial equation as

log(x^2- x) = 3

Remember that logarithms are simply another way of writing exponents. The logarithm  is just another way of writing the fact . Keeping that in mind, we can express our logarithm in terms of exponents as

log(x^2- x) = 3 --> x^2- 2x = 8

2³ = 8, so we can replace the left side of our equation with 8 to get

x^2 - 2x - 8 =0

(x-4)(x+2) = 0

Hence x = 4 0r -2. As x cannot be negative so it would be 4.

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Answer

16/9

Step-by-step explanation:

I don't know if this is right or not and I'm sry if its wrong

but 16/9 divided by 4 = 9/4 

Answer:

v =L× b×h

=5× 4×3

=60 cubic centimeter

The statement which is true is the slopes are all positive in quadrant I.

Given the differential equation is dy/dx=2x-2y

A differential equation is an equation that contains at least one derivative of an unknown function, either a normal differential equation or a partial differential equation.

Given dy/dx=2x-2y

now slope=2x-2y

Along x-axis, y=0.

So, slope=2x≠0.

Since it depends upon x hence the slope along the y-axis are not horizontal.

Along y-axis, x=0.

So, slope=-2y≠0.

Since the slope along the x-axis are also not horizontal.

In quadrant I

x,y≥0

So, dy/dx≥0

Hence the slopes are all positive in quadrant I.

In quadrant IV,

x≥0,y≤0

so, dy/dx is  not always positive.

This, the slope are not all positive in quadrant IV.

Hence, the slope are all positive in quadrant I for the differential equation dy/dx=2x-2y.

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The solution to the given fraction 7/11 = x/9 to the nearest tenth is 5.7

7/11 = x/9

7 × 9 = 11 × x

63 = 11x

x = 63/11

x = 5.72727272727272

Approximately,

x = 5.7

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