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2 years ago

Functions and linear

Mathematics

1 answer:

Westkost [7]2 years ago

8 0

Answer:

m= 1x or x and b=-2

Graph it and you can find out more

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Find the volume of the solid<br><br><br><br> The problem is in the picture!!!!!!!!!

Anna35 [415]

The answer will be C. In the picture you will see 2144 which is the answer.

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3 years ago

Find the area of the figure. 7 cm 4 cm 4 cm 1 10 cm 10 cm 8 cm 19 cm

Paladinen [302]

Answer:

132cm^2

Step-by-step explanation:

first break the shape up into component shapes. the shape is made of two rectangles and two triangles. the area for a rectangle is A=BH where b is base and h is height. the area of a right triangle is BH/2 where b is the base and h is the height. you are given the base and height of the top rectangle 4x7. To find the area of the triangles and the second rectangle you have to do some algebra. the total base length is 19cm and the top rectangle has a length of 7cm so 19-7 leaves you with 12cm to spare and there is two symmetric triangles on each side so 12/2 is 6. now we know all the measurements and can solve.

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2 years ago

The area of a rectangular wall of a barn is 32 square feet. It’s length is 4 feet longer than the width. Find the length and wid

Gnom [1K]

Answer:

Step-by-step explanation:

Givens

Area = L * w

w = x

l = x + 4

Area = 32

Solution

32 = x(x + 4)

32 = x^2 + 4x

x^2 + 4x - 32

(x + 8)(x - 4)

Only x - 4 is going to make any sense

x - 4 = 0

x = 4

The width is 4

The length is 4 + 4 = 8

4*8 = 32 which is the area.

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3 years ago

44 lb 6 oz = ____ ounces

hodyreva [135]

Answer:

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Step-by-step explanation:

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2 years ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

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3 years ago