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3 years ago

Please help I have 9 min

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

I believe the answer is D because the pathagreom theorum proves it true (excuse my terrible grammar)

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HELPPPPPPP GIVING 100 POINTS FOR ANSWER.......Ivan bought a box of macadamia nuts to bring home from his trip. A rectangular pri

Alex777 [14]

Answer:

18 x 15 x 3 is 810. The volume is 810.

Step-by-step explanation:

Volume of cuboid= Length*Width*Breadth

Volume of container= 18 cm X 15 cm X 3cm

Volume of container=810

Hope i helped

4 0

3 years ago

Read 2 more answers

Which relation is a function?

lina2011 [118]

Answer:

Step-by-step explanation:

For a relationship to be a function

Each value of x in the domain can only have 1 unique value of y in the range. That is, one-to-one correspondence.

The only relation which meets this criteria is B

7 0

3 years ago

Read 2 more answers

Which expressions are equivalent to 12x - 6?

iren2701 [21]

<h3>Answer</h3>

B and E

<h3>Explanation</h3>

A) It's wrong because

$- 6 \times (2x - 1) = - 12x + 6$

B) It's right because

$6 \times (2x - 1) = 12x - 6$

C) It's wrong because

$6x \times (2 - 1) = 6x \times 1 = 6x$

D) It's wrong because

$- 6x \times (2x - 1) = - 12 {x}^{2} + 6$

E) It's right because

$- 6 \times ( - 2x + 1) = 12x - 6$

F) It's wrong because

$6 \times ( - 2x + 1) = - 12x + 6$

8 0

3 years ago

2 2/10 x 6 2/5 please help me!!!!! Thank you!!!

evablogger [386]

The answer is 24/25.

8 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

3 years ago