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3 years ago

What is the missing side length?

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

7 0

In a 45-45-90 triangle, the legs are equal and the hypotenuse is sqrt2 times longer than either leg

y (leg) = 5 rt 2 / 2

x (hyp) = 5 rt 2 / 2 x rt 2
= 5 * rt 2 * rt 2 / 2
= 5 * 2 / 2
= 10/2
x = 5

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Factor the expression. 36y2 – 84y – 147

Lubov Fominskaja [6]

Answer:

1) 3 (6y+7) (2y−7)

2) 12y2+28y−49

3) 36y2+168y+147

4) 36y2−84y−147

5) 3(6y+7)(2y−7)

8 0

4 years ago

To find the quotient of 3 divided by one-sixtto find the quotient three-fourths divided by one-eighth,h, multiply 3 by

ollegr [7]

The quotient of 3 by 1/ 6 and that of 3/4 by 1/ 8 is 1/2 and 4 respectively

<h3>How to find the quotient</h3>

From the given equation, we have to find

= 3/4 ÷ 1/8

First , we take the denominator of the second fraction to be the numerator, we have

$\frac{3}{4} \times \frac{8}{1}$

Multiply through

$\frac{24}{4}$

$6$

To find the quotient of 3 by 1/6

$3 \times \frac{1}{6}$

$\frac{3}{6}$

$\frac{1}{2}$

Thus, the quotient of 3 by 1/ 6 and that of 3/4 by 1/ 8 is 1/2 and 4 respectively

Learn more about quotient here:

brainly.com/question/629998

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4 0

2 years ago

What is the diameter of the smallest circle that would enclose 2,000 pennies?

OlgaM077 [116]

Answer: 84.9

Step-by-step explanation:

Model's equation: y=1.11x^2

Since y = 2000

y = 1.11x^2

2000 = 1.11x^2

x^2 = 2000/1.11

x^2 = 1801.8

x = 42.45

Diameter of the smallest circle will be:

= 2x

= 2 × 42.45

= 84.9

7 0

3 years ago

Sylvie needs $110 for a concert ticket. She already has $16 and she can earn the rest by working 10 hours at her job. If h repre

antoniya [11.8K]

A and c is the correct answers

7 0

3 years ago

I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

<u>For the question a *</u> you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

<u>For question b, do the same procedure</u>.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

<u>Finally for the question c we have</u>

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

8 0

3 years ago