50 Points! A section of wall for a mural is shaped like a parallelogram with a base length of 7 feet and a height of 11 feet. A can of paint will cover 20 square feet.

nvm it was 4

Answer:

z=3t

Step-by-step explanation:

multiply it by 3

.................................................................................

Answer:

D) none of the above

Step-by-step explanation:

16+20+30+30=96

96/4 =24

Answer:

A). thickness are correct

Answer:

the answers 1, 4, and 5.

m∠N = 59° and m∠O = 121°.

Option 3 and 4 are correct.

A parallelogram is a geometric object with sides that are parallel to one another in two dimensions. It is a form of polygon with four sides (sometimes known as a quadrilateral) in which each parallel pair of sides have the same length. A parallelogram has adjacent angles that add up to 180 degrees. You must have studied a variety of 2D shapes and sizes in geometry, including circles, squares, rectangles, rhombuses, etc. Each of these forms has a unique set of characteristics.

As per the given data:

LMNO is a parallelogram

The sum of adjacent angles in a parallelogram is always equal to 180°

11x + 6x - 7 = 180

17x = 187

x = 11

m∠L = m∠N {opposite angles in a parallelogram are equal}

m∠N = 6x - 7 = 6(11) - 7 m∠N

m∠L = 59°

m∠M = m∠O {opposite angles in a parallelogram are equal}

m∠M = 11x = 11(11) = 121°

m∠O = 121°

Hence, m∠N = 59° and m∠O = 121°.

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Let's see

LHS

Keep it stay

RHS

Use distributive law

Hence proved

angle ksp+61=108(being straight angle)

or,angle ksp=180-61

angle ksp=119

How to determine the total amount of rainfall

From the dot plot, we have the following frequency table

Rainfall    Frequency

0                  4

1/8               5

1/4               1

3/8              2

1/2               1

1 1/4             1

= 3/8 inches

Answer:

Step-by-step explanation:

[tex]=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}x\sec^2(xy)dy\\=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}\sec^2(xy)d(xy)\\=5\displaystyle\int_{0}^{\pi/6}dx\tan(xy)|_{y=0}^{y=2}[/tex]

[tex]=5\displaystyle\int_{0}^{\pi/6}\tan(2x)dx\\=-\frac{5}{2}\ln\cos(2x)|_{0}^{\pi/6}\\=-\frac{5}{2}[\ln\cos(\pi/3) - \ln\cos(0)]\\[/tex]

[tex]=-\frac{5}{2}\ln{\frac{1}{2}[/tex]

Answer:

24.

Step-by-step explanation:

Solution,

By Prime Factorisation method,

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

576 = (2)² × (2)² × (2)² × (3)²

Now,

By Taking square roots on both the sides,

√576 = √ (2)² × (2)² × (2)² × (3)²

√576 = 2 × 2 × 2 × 3

√576 = 24.

Answer:

97.16

Step-by-step explanation:

in total, you have $668.00.

To reserve enough for the last day, you need to subtract the last day's budget:

This means that, for the remaining 6 days in the week, you have $583.00.

Assuming an equal amount is being spent each day, you can calculate the daily budget by dividing the remaining total by the number of days left.

To avoid going over budget, you can spend $97.16 per day for the other six days.

Answer:

18 grams,160 grams,2,100 grams,2 kilograms,17 kilograms,208 kilograms.

The measure from least to greatest is 18 gm < 10 gm< 17 Kg < 2Kg < 2300 gm < 208 Kg.

The arrangement of numbers or other things in increasing order from least to greatest is known as ascending order. Ascending order is demonstrated by numbers on a number line, which are listed from left to right.

Given:

18gm, 17 Kg, 100 gm, 2 Kg, 2300 gm, 208 Kg

Converting all the units into gram we have

18 , 1700,  100, 2000, 2300, 208000

So, the mass from least to greatest.

18 < 100 < 1700< 2000< 2300 < 208000

Hence, 18 gm < 10 gm< 17 Kg < 2Kg < 2300 gm < 208 Kg.

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[tex]\dfrac 6p = x+a\\\\\implies \dfrac 6p \times\dfrac 16 = (x+a) \times\dfrac 16 ~~~~~~;\left[\text{Multiply both sides by}~ \dfrac 16 \right]\\\\\implies \dfrac 1p = \dfrac{x+a}6\\\\\implies p= \dfrac{6}{x+a}~~~~~~~~~~~~~~~~~~~~;[\text{Cross multiply}][/tex]

Step-by-step explanation:

63/37 = 1.702 = 1.7

#Ihopethishleps

Answer:

The graph should look almost exactly like this. I used a virtual graphing chart to assist, since it is capable of showing more than paper alone.

Answer:

Here is your answer.

Step-by-step explanation:

Just have a good day

The dimension of the unknown variable x of the deck is equal to 14 feet in length.

A rectangle is a closed 2-D shape, having 4 sides, 4 corners, and 4 right angles (90°).The opposite sides of a rectangle are equal and parallel.

Given in the question is a diagram showing a plan for deck. From the data given, we can write -

Area of Deck = 511 ft.

The value of x can be found out by subtracting the area of the deck from the  area of rectangle having dimensions of length 29 ft and width (16 + 4) = 20 ft. The remaining area will be equated to the sum of the area of rectangle at top right corner having dimensions of length 9 ft and width 4 ft and the area of triangle at lower right corner.

Firstly, we will find the dimensions of the triangle at the lower right. The height of the triangle will be = (20 - 9) ft = 11 ft. The base of the triangle will be = 29 - ( x + 9) ft = (20 - x) ft. Now -

[Area of Rectangle (29 x 20)] - [Area of Deck] = [Area of Rectangle (9 x 4) + Area of triangle (lower right)].

Mathematically -

580 - 511 = 9 x 4 + 1/2 × (20 - x) × 11 = 69

36 + 5.5(20 - x) = 69

36 + 110 - 5.5x = 69

146 - 69 = 5.5x

5.5x = 77

x = 14 ft

Therefore, the dimension of the unknown variable x of the deck is equal to 14 feet.

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[tex] \Large \bold{SOLUTION\ 1:} [/tex]

[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]

[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]

[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

[tex] \Large \bold{SOLUTION\ 2:} [/tex]

[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]

[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

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Answer:

Step-by-step explanation:

In a regular pentagon, all the angles and sides are equal. So, to find an angle, divide 540 by 5.

Sum of all angles of regular pentagon = 540

measurement of one angle = 540 ÷ 5

                                             = 108°

∠N = 108°

Answer:

B

Step-by-step explanation:

Answer:

B and E

Step-by-step explanation:

Answer:

15 :)

Step-by-step explanation:

brainliest please

Answer:

The answer is -125

Proof:

In the given histogram, the correct option regarding the representation of the fourth bin is:

at least 12 pounds but less than 16 pounds.

An histogram is a graph that shows the number of times each element of x was observed. The lower bound is inclusive, while the upper bound is exclusive.

Hence, considering the inclusiveness and exclusiveness of the bounds, the correct option regarding the fourth bin, given by [12, 16) is:

at least 12 pounds but less than 16 pounds.

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Answer:

The answer is "at least 12 pounds but less than 16 pounds"

Explanation:

I took the quiz! Hope this helps! :)

Answer: 48.6°

Step-by-step explanation:

Angle of elevation = sin inverse (6/8)

= 48.6 degrees

Therefore, Option C is right answer

Using Chebyshev's Theorem, considering a standard deviation of 40, we have that at least 75% of the students scored between 360 and 520.

When we have no information about the population distribution, Chebyshev's Theorem is used. It states that:

In this problem, considering a standard deviation of 40, we have that:

440 - 2 x 40 = 360.

440 + 2 x 30 = 520.

Within 2 standard deviations of the mean, no information about the distribution, hence, at least 75% of the students scored between 360 and 520.

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Answer:

x = 5

Step-by-step explanation:

3^2 + 4^2 = x^2

9 + 16 = x^2

25

Now we need to find the square root of 25 :)

The answer is 5

X = 5

Have a great day!!

Please rate and mark brainliest!!

Answer:

When it's negative when it goes to the other side it becomes a positive.....

Y=(x-4)(x-1)

Make y 0 when your calculating for x

Make x 0 when your calculating for y

0=(x-4)(x-1)

Therefore x=4 or x=1

Answer:

Step-by-step explanation:

The equation factors into

(x^2 + 14x + 49) =16                      Factor

(x + 7)^2 = 16                                 Take the square root of both sides

√(x + 7)^2 = √16

x + 7 = +/- 4

x + 7 = + 4                                      Subtract 7 from both sides

x  = 4 - 7

x = - 3

x + 7 = - 4                                      Subtract 7 from both sides.

x = - 4 - 7

x = - 11

x1 = -11

x2 = - 3

Check the picture below.

[tex]\textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(255^o)\implies sin(135^o+120^o) \\\\\\ sin(135^o)cos(120^o)~~ + ~~cos(135^o)sin(120^o) \\\\\\ \left( \cfrac{\sqrt{2}}{2} \right)\left( -\cfrac{1}{2} \right)~~ + ~~\left( -\cfrac{\sqrt{2}}{2} \right)\left( \cfrac{\sqrt{3}}{2} \right)\implies -\cfrac{\sqrt{2}}{4}~~ - ~~\cfrac{\sqrt{6}}{4}\implies \cfrac{-2-\sqrt{6}}{4}[/tex]

The amount of oxygen in the flask through the intake tube illustrates proportions

The flask 99.35% will contain of oxygen after 5L have passed through the intake tube

To do this, we make use of the following representations

So. we have the following derivative

dP = dV - P * dV

Divide through by dV

dP/dV  = 1 - P

Next, we set up the integral

[tex]\int\limits^x_{0.040} \frac {1}{1 - P}dP= \int\limits^5_0 {dV}[/tex]

Integrate both sides

[tex]-\ln(|P - 1|)|\limits^x_{0.04} = 5[/tex]

Expand

-ln(|x - 1|) + ln(|0.04 - 1|) = 5

Evaluate the difference

-ln(|x - 1|) + ln(|-0.96|) = 5

Express as a fraction

ln(0.96/|x - 1|) = 5

Express both sides as exponents with a base of 2

e^ln(0.96/|x - 1|) = e^5

Solve for x

x = 1 - 0.96/e^5

Evaluate the difference

x = 0.9935

Express as percentage

x = 99.35%

Hence, the flask 99.35% will contain of oxygen after 5L have passed through the intake tube

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Answer:

Step-by-step explanation:

7 hours and 37 minuties

Answer:

5 hours

Step-by-step explanation:

it lasted 5 hours for the tournament

Step-by-step explanation:

the inverse of 2x-3 is

[tex] \frac{1}{2x - 3} [/tex]

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