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3 years ago

33) 90÷18-18÷6-6÷3= Preciso de ajudaaaa

Mathematics

1 answer:

AleksandrR [38]3 years ago

3 0

Answer:

Step-by-step explanation:

(90÷18)-(18÷6)-(6÷3)= 72-3-2= 67

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What is the sum of 8 and 7, doubled

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(8+7) x 2

= 15 x 2

= 30

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3 years ago

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What is 1/6+7/10 as a equivalent fraction

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Answer: 13/15

Step-by-step explanation:

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4 years ago

Solve (16×^3-3×^2+43×-5)÷(×^2-2)

Oksana_A [137]

Answer:

3872/(x^2 - 2)

Step-by-step explanation:

Simplify the following:

(16^3 - 3^2 - 5×43)/(x^2 - 2)

3^2 = 9:

(16^3 - 9 - 5×43)/(x^2 - 2)

16^3 = 16×16^2:

(16×16^2 - 9 - 5×43)/(x^2 - 2)

| 1 | 6

× | 1 | 6

| 9 | 6

1 | 6 | 0

2 | 5 | 6:

(16×256 - 9 - 5×43)/(x^2 - 2)

16×256 = 4096:

(4096 - 9 - 5×43)/(x^2 - 2)

43 (-5) = -215:

(4096 - 9 + -215)/(x^2 - 2)

4096 - 9 - 215 = 3872:

Answer: 3872/(x^2 - 2)

7 0

3 years ago

Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank i

shepuryov [24]

Answer:D) 1 hr

Step-by-step explanation:

Given

Pump A and B can fill tank in $\frac{6}{5} hr$

Pump A and C can fill tank in $\frac{3}{2} hr$

Pump B and C can fill tank in $2 hr$

Let A be the total hr taken A therefore rate of $A=\frac{1}{A} /hr$

Let B be the total hr taken B therefore rate of $B=\frac{1}{B} /hr$

Let C be the total hr taken C therefore rate of $C=\frac{1}{C} /hr$

$\frac{1}{A}+\frac{1}{B}=\frac{5}{6}$ -----1

$\frac{1}{B}+\frac{1}{C}=\frac{1}{2}$ -----2

$\frac{1}{A}+\frac{1}{C}=\frac{2}{3}$ -----3

Adding 1,2 & 3

$2(\frac{1}{A}+\frac{1}{B}+\frac{1}{C})=\frac{5}{6}+\frac{1}{2}+\frac{2}{3}$

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{2}$

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1$

thus time taken by A,B and C combined is 1 hr

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3 years ago

Find the unknown side length

Westkost [7]

For any right triangle, we can use the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle, the legs when squared and added together will be equal to the hypotenuse squared.

In mathematical notation: $a^2 + b^2 = c^2$

Where the variables a and b are the legs and the variable c is the hypotenuse.

Because we know the two side lengths of the triangle, we can solve for the unknown side.

We know the length of one of the legs and the hypotenuse.

Plug in the values.

$10^2 + b^2 = 24^2 \\ \\ 
100 + b^2 = 576 \\ \\ 
b^2 = 476 \\ \\ 
$

$b = \sqrt{476}$

So, the square root of 476 is the unknown length.

3 0

3 years ago