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3 years ago

What is the length of the diagonal (d) is the solid below

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

8 0

420!!!!!!!!!!!!!!!!!!

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Becky is responsible for purchasing a flower centerpiece for an event. The nearby florist shop offers 20 types of flowers, 13 di

Soloha48 [4]

Multiply all the numbers together to get the total amount of possible outcomes

20 x 13 x 35 = 9100

There are 9100 possible outcomes that Becky can get

hope this helps

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3 years ago

If Jenny has 20 donuts and she gives away 10 how many does she have left?

stiks02 [169]

10 donuts to be exact

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3 years ago

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6x3 + 8x2 – 2x + 4 and 10x3 + x2 + 11x + 9

balu736 [363]

6x3+8x2-2x+4=30-2x

10x3+x2+11x+9=21+13x

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3 years ago

(1) The Yearling is one of my favorite novels.(2) It is by Marjorie Kinnan Rawlings. (3) It is about a yearling, which is a youn

weqwewe [10]

Yea u have to do that the most effective is the 10 bc penny is right

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3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago