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3 years ago

Can anyone help me please

Mathematics

1 answer:

Nady [450]3 years ago

6 0

Yo sup??

since WY is the perpendicular bisector we can say

3x-5=2x+3

x=8

Hope this helps

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25 − (8 − 4) ÷ 6 • 3=what plzzzz answer!<br> Will mark Brainliest!!!!

serg [7]

25 - (8-4) /6 *3 = 25 - 2 / 6 * 3
=25-1
=24 #

“/”= divide
“*”= multiply

4 0

3 years ago

Read 2 more answers

What is the interquartile range of the data set?

Andrej [43]

Answer:

Should be 10 if there is nothing for th stem value of 3.

Step-by-step explanation:

The stem is the tens place, and the leaf is the ones. You would get 28 40 43 43 45 50 50 for your numbers. If you take Q3-Q1, you get your answer, 10.

7 0

3 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

6y^2-25y+14 factor <br><br> Help please

LUCKY_DIMON [66]

Answer:

(3y - 2)(2y - 7)

Step-by-step explanation:

To factorise the quadratic

Consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term

Product = 6 × 14 = 84 and sum = - 25

The required factors are - 4 and - 21

Use these factors to split the y- term

6y² - 4y - 21y + 14 ( factor the first/second and third/fourth terms )

2y(3y - 2) - 7(3y - 2) ← factor out (3y - 2) from each term

= (3y - 2)(2y - 7) ← in factored form

8 0

3 years ago

List the following fractions from least to greatest 1/7 6/7 4/7 3/7

neonofarm [45]

Answer:

1/7, 3/7, 4/7, 6/7

Step-by-step explanation:

There's like denominators(bottom number) so you would list the numerators(top number) from least to greatest.

5 0

3 years ago