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mestny [16]
3 years ago
8

PLEASE HELP!!! FIND X IN THE DIAGRAM PROVIDED!! WILL GIVE BRIAIN!!

Mathematics
2 answers:
Softa [21]3 years ago
7 0
D is your answer. Hi
iren2701 [21]3 years ago
6 0

:\implies \sf 3x⁰+45⁰+90⁰ = 180⁰

:\implies \sf 3x⁰+135⁰ = 180⁰

:\implies \sf 3x⁰ = 45⁰

:\implies \sf x = \dfrac{45}{3}

:\implies \sf x = 15⁰

hence, last option is right

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A square coaster has a side length of 512 feet on each side.
Mariana [72]
Hi Mitsu there is no fraction to this. You multiply the base and height to get the area. So 512 x 512 = 262144
5 0
2 years ago
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One spinner has three equally sized sectors numbered 1, 2, and 3. A second spinner has two equally sized sectors labeled Top (T)
SSSSS [86.1K]
I would say, well, if there are 3 sides that are equal on spinner 1. They are equal. But, we are focusing on spinner two that has two sides. If there is only 2 equal sides then, they would have to be straight across from each other. So that is 180. 360 is full circle. take away 180 to get answer. 
360-180=180
I hope this helped. And sorry if it is not the correct answer. But that is how I would of answered it
4 0
3 years ago
Read 2 more answers
Solve the following equation:<br> V= 2(ab+bc+ca), for a
Nadya [2.5K]

V = 2ab + 2ca + 2bc

2ab + 2ca = V - 2bc

2a * b + c = V - 2bc

Answer: a = v - 2bc / 2 * b + c



Hope that helps!!!!! : )

8 0
4 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

6 0
4 years ago
I need help with this practice I am having a tough time solving it properly
Monica [59]

Given:

There are given that the parent functions as a cosine function:

Where,

The amplitude of the function is 9.

The vertical shift is 11 units down.

Explanation:

To find the cosine function, we need to see the standard form of the cosine function:

f(x)=acos(bx+c)+d

Where,

a is the amplitude of the function,

Now,

According to the question:

The amplitude of the function is 9, which means:

f(x)=9cos(bx+c)+d

The vertical shift is 11 units down, which means:

f(x)=9cos(bx+c)-11

For period:

\begin{gathered} f(x)=-9cos(\frac{12\pi}{7}x+0)-11 \\ f(x)=-9cos(\frac{12\pi}{7}x)-11 \end{gathered}

Final answer:

Hence, the cosine function is shown below;

f(x)=-9cos(\frac{12\pi}{7}x)-11

4 0
1 year ago
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