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3 years ago

Help me answer please

Mathematics

1 answer:

fiasKO [112]3 years ago

6 0

Answer:

Step-by-step explanation:

b=4x+1 h=x+1 A=175

A=bh

Substitute: 175=(4x+1)(x+1)

Expand: 175=4x^2+5x+1

Subtract 175

4x^2+5x-174=0

Solve w/quadratic formula

(-5 <u>+</u> sqrt(25+2784))/8

Simplify: x=6 or -29/4

Distance is positive, so x=6

4x+1=4*6+1=25

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Can someone SMART actually help me

lisov135 [29]

The answer is 196/3 pi in^3. All you have to do is multiply 7x7, then you have to multiply that by 1/3 and 4, and you get 65.333333333. Then, you have to turn it into a fraction. You have to multiply 65 by 3, then add one, and you get 196/3 pi in^3.

7 0

3 years ago

if 5.25 MB have been downloaded and 2.25 MB are left. what percent of the application has already been downloaded?

ollegr [7]

5.25 + 2.25 = 7.50 MB

x/100 x (7.50) = 5.25
7.50x/100 = 5.25
7.50x = 5.25(100)
7.50x = 525
x = 525/7.50
x = 70
Hence, 70% of the application has already been downloaded.

6 0

3 years ago

Can somebody please help me out

Alekssandra [29.7K]

Answer:

with what????

8 0

3 years ago

A cylinder has a height h, and a radius (3h-2)

tekilochka [14]

Answer:

$V= 4\pi (9h^2 -12h + 4)$

Step-by-step explanation:

Given

Shape: Cylinder

$Radius = 3h - 2$

$Height = h$

$Base\ Area = \pi (3h - 2)^2$

Required

Simplify the volume

The volume is calculated as:

$Volume = Base\ Area * Height$

Substitute values for Base Area and Height

$V= \pi(3h-2)^2*h$

Expand the bracket

$V= \pi(3h-2)(3h-2)*h$

Open brackets

$V= \pi (9h^2 -6h - 6h + 4) *h$

$V= \pi (9h^2 -12h + 4) *h$

$V= 4\pi (9h^2 -12h + 4)$

6 0

2 years ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$

∴

$\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}$

8 0

3 years ago