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3 years ago

6

Which set of numbers represents the lengths of the sides of a right triangle? 1. {7, 24, 25} 2. {9, 16, 23} 3. {10, 12, 14} 4. {

14, 16, 18}?

1 answer:

wolverine [178]3 years ago

5 0

1
${7}^{2} + {24}^{2} = {25}^{2}$

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Misty wishes to obtain 85 ounces of a 40% acid solution by combining a 72% acid solution with a 25% acid solution. How much of e

Y_Kistochka [10]

13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed

<u>Step-by-step explanation:</u>

Total mass of acid required= 85 ounces

Let the mass of 72% acid be 'a'

Let the mass of 25% acid be 'b'

a + b = 85

b = 85-a

85(40/100) = a(72/100) + b(25/100)

85(2/5) = a(72/100) +(85- a) (25/100)

34 = (72a/100) + (2125/100) - (25a/100)

34 - (2125/100) = (72a + 25a) /100

(3400-2125)/100 = 97a /100

97a = 1275

a = 13.15 ounces

b = 85 - 13.14

b = 71.85 ounces

13.15 ounces of 72% acid and 71.85 ounces of 25% acid are needed

4 0

3 years ago

If f(x) = 2x^3– 5x+ 2, then what is the remainder when f(x) is divided by<br> X – 4?

Mashutka [201]

The remainder is 110 when $f(x) = 2x^3-5x+ 2$ is divided by x-4

Step-by-step explanation:

We need to find the remainder when $f(x) = 2x^3-5x+ 2$ is divided by x-4

Solving:

The solution is attached in the figure below.

The quotient is: $2x^2+8x+27$

The remainder is: 110

Keywords: quotient, division

Learn more about Division and quotient at:

#learnwithBrainly

5 0

3 years ago

6a - 5b when a = -3 and b = 4

DIA [1.3K]

Given.

6a - 5b

Plug in values.

6(-3) - 5(4) =

-18 - 20 =

-38

8 0

3 years ago

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Plsplspls help!!!

RideAnS [48]

Answer:

a) the answer if the substitute is 300

7 0

2 years ago

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago