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3 years ago

I tried to answer this but not getting the right answer. Can someone help me plzzz

Mathematics

1 answer:

katovenus [111]3 years ago

8 0

Answer: 17

Step-by-step explanation:

Use the Pythagorean theorem.

C^2 = A^2 + B^2

So, C is always the hypotenuse which is the longest side length and the side length diagonal of the right angle. Side "s" is the hypotenuse.

Square sides AB and BD

It looks like this:

C^2 = 8^2 + 15^2

8^2 = 64

15^2 = 225

225 + 64 = 289

the square root of 289 is 17

17 = s

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What situation should be used to rewrite 16(x^3+1)^2 -22(x^3+1)-3=0 as a quadratic equation

Oliga [24]

$16(x^3+1)^2-22(x^3+1)-3=0$

Use substitution: $x^3+1=t$

$16t^2-22t-3=0\\\\16t^2-24t+2t-3=0\\\\8t(2t-3)+1(2t-3)=0\\\\(2t-3)(8t+1)=0\iff2t-3=0\ \vee\ 8t+1=0$

$2t-3=0\ \ \ |+3\\\\2t=3\ \ \ |:2\\\\t=\dfrac{3}{2}\\..............................\\8t+1=0\ \ \ \ |-1\\\\8t=-1\ \ \ \ |:8\\\\t=-\dfrac{1}{8}$

we're going back to substitution:

$x^3+1=\dfrac{3}{2}\ \vee\ x^3+1=-\dfrac{1}{8}\ \ \ \ |subtract\ 1\ from\ both\ sides\ of\ the\ equations\\\\x^3=\dfrac{1}{2}\ \vee\ x^3=-\dfrac{9}{8}\\\\x=\sqrt[3]{\dfrac{1}{2}}\ \vee\ x=\sqrt[3]{-\dfrac{9}{8}}$

$x=\dfrac{1}{\sqrt[3]2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}\\\\\boxed{x=\dfrac{\sqrt[3]4}{2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}}$

4 0

3 years ago

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Gnoma [55]

Answer:

x2 +1

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Together, Linda and Penny sold 492 boxes of Girl Scout cookies. Penny

boyakko [2]

Answer:

222

Step-by-step explanation:

Let L = number of boxes Linda sold.

Let P = number of boxes Penny sold.

The sold a total of 492 boxes.

L + P = 492

Penny sold 48 more boxes than Linda.

P = L + 48

Substitute P with L + 48 in the first equation and solve for L.

L + L + 48 = 492

2L + 48 = 492

2L = 444

L = 222

Answer: Linda sold 222 boxes.

Check:

Penny sold 48 more boxes, so she sold 222 + 48 = 270.

Add the numbers they sold: 222 + 270 = 492

The difference is 48, and the total is 492. Our answer is correct.

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3 years ago

What is the IQR (interquartile range) of the data set? 33, 37, 45, 61, 64, 67

Nataly_w [17]

solution is first os all find median

median is( 45+61)/2=53

find median of no in left of 53

it is 37

find median of no in right of 53

it is 64

now interquartile range=64-53=27

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4 years ago

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Zepler [3.9K]

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I divided the worth of clothe and the tax

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