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trasher [3.6K]
3 years ago
12

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

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4 0
3 years ago
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Which relation is a function?
lina2011 [118]

Answer:

B

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
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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
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