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3 years ago

Thirty and 25 hundredths in decimal form

1 answer:

4 0

30.25. Thirty is the whole number, therefore it goes before the decimal. The 25 hundredths come after the decimal. The first Slot after the decimal is the tenths. And the second is the hundredths. .25 is twenty five hundredths.

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3(-4/7) I don’t know how to get the answer

WARRIOR [948]

Answer:

Step-by-step explanation:

if its 3 - 4/7 then it = 2 3/7

if its 3*-4/7 then it = -12/7 = - 1 5/7

7 0

4 years ago

Can you plz help me with this problem I’ve been stuck on it for a long time

Tpy6a [65]

I think the answer is B

6 0

3 years ago

Read 2 more answers

In order to make an A on her project, Sarah needs at least 160 points. She has already turned in part of her work and has been g

antoniya [11.8K]

to get an A, she needs >=160 points and she already has 125 points

let 'x' be the number of points on the last part of her project then

125 + x >= 160

x >= 160 - 125

x >= 35

Sarah needs at least 35 points on the last part of her project.

8 0

3 years ago

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Write the linear equation from the given information in slope intercept form (-1,3); slope=-3

nadya68 [22]

Answer:

Step-by-step explanation:

The standard expression of equation of a line in slope-intercept form is expressed as;

y = mx+c

m is the slope

c is the intercept

Given

slope m = -3

Point (x, y) = (-1, 3)

x = -1 and y = 3

Get the intercept

To get the intercept c, we will substitute the given values into the equation above to have;

y = mx+c

3 = -3(-1)+c

3 = 3 + c

c = 3-3

c = 0

Substitute m = -3 and c = 0 into the equation above;

y = mx+c

y = -3x+0

y = -3x

Hence the required linear equation is y = -3x

4 0

3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago