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sergiy2304 [10]

2 years ago

6

Harriet would like to frame her graduation picture. The picture is 8 inches wide and 11 inches long. She wants the total area of

the picture and frame to be 108 square inches. How wide should the border around the picture be?

1 answer:

LekaFEV [45]2 years ago

8 0

Answer: i explained

Step-by-step explanation: Just do

8 times 11 = *your answer*

then

*your answer* times/subtract/divided by 108

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Subtract. (4x2+8x−2)−(2x2−4x+3) Enter your answer, in standard form, in the box.

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the answer is 2x2+12z-5 (its also in the picture)

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What is the equation of the graph below?

kherson [118]

Answer:

A. y = sine (x + 90 degrees)

Step-by-step explanation:

y = cosine(x) is a curve that crosses the y-axis at y = 1 and completes one cycle at 360 degrees.

sine(x) have the same curve than cosine(x), but translated 90° to the right respect cosine(x)

f(x + c) translates f(x) horizontally c units to the left.

Then, sine(x + 90) is equivalent to cosine(x)

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2 years ago

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You are facing North. Turn 90 degrees left. Turn 180 degrees right. Turn around to reverse your direction. Turn 45 degrees left.

andriy [413]

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You are facing north. 90 degrees left is west. 180 degrees is now east. Reversing your direction is the same as 180 so your now back to facing west. Turn 45 left you’re now facing south. Reverse again and you’re facing north.

5 0

2 years ago

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Which are the solutions of x2 = –11x + 4? StartFraction negative 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFr

Marysya12 [62]

Answer:

$x_1=-\frac{11}{2}-\frac{\sqrt{137} }{2}\\\\x_2=-\frac{11}{2}+\frac{\sqrt{137} }{2}$

Step-by-step explanation:

Given the following quadratic equation:

$x^2 = -11x + 4$

The steps to solve it are:

1. Move the terms to one side of the equation:

$x^2+11x- 4=0$

2. Apply the Quadratic formula $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ .

In this case we can identify that:

$a=1\\b=11\\c=-4$

Then, substituting these values into the Quadratic formula we get the following solutions:

$x=\frac{-11\±\sqrt{11^2-4(1)(-4)} }{2(1)}$

$x_1=-\frac{11}{2}-\frac{\sqrt{137} }{2}\\\\x_2=-\frac{11}{2}+\frac{\sqrt{137} }{2}$

3 0

2 years ago

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Given △DEF, which is not equal to cos(F)? sin(F). sin(D). tan(F). cos(D)

valentina_108 [34]

Answer: The correct option is tan(F).

Explanation:

In the given figure the triangle is a right angle triangle because $\angle E=90^{\circ}$ .

It is also given that $DE=EF=5$ and $DF=5\sqrt{2}$ .

Since it is an isosceles right angle triangle, therefore the value of perpendicular and base is same for both angles D and F, which is 5.

$\cos (F)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (F)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (D)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(D) is $\frac{1}{\sqrt{2}}$ .

$\tan (F)=\frac{Perpendicular}{Base}=\frac{5}{5} =1$

The value of tan(F) is 1. Which is not equal to $\frac{1}{\sqrt{2}}$ .

$\cos (D)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(D) is $\frac{1}{\sqrt{2}}$ .

Therefore, the value of tan(F) is not equal to the value of cos(F), so the correct option is third.

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2 years ago

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