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3 years ago

Find the missing numbers then write each as a decimal 3/15=1/?=?/10

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

3 0

3/15=1/5=2/10 and each one is equal to 0.2, I hope this helps cx

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Jake plans to use a ramp to make it easier to move a piano out of the back of his truck. The back of the truck is 83 centimeters

storchak [24]

Answer:

134.4 centimeters

Step-by-step explanation:

This question is asking us for a missing leg on the triangle, lets call it x. Using the Pythagorean theorem (a²+b²=c²), we can figure out what value is missing (either a or b, they are interchangeable).

replace the values in the equation.

83²+x²=158²

get rid of the exponents.

6,889+ x² = 24,964.

subtract 6,889 from 24,964.

x²=18,075.

find the square root on both sides of the equation.

x=134.4432966.

I am assuming you want this rounded to the nearest tenth.

134.4 is your answer.

Read more on Brainly.com - brainly.com/question/14113423#readmore

8 0

3 years ago

^^^^^^^^^^^^^^^^^^^^

telo118 [61]

ANSWER

The correct answer is C

EXPLANATION

We want to find the quotient:

$- \frac{10}{19} \div ( - \frac{5}{7} )$

We multiply by the reciprocal of the second fraction:

$- \frac{10}{19} \times ( - \frac{7}{5} )$

We cancel out the common factors to obtain:

$- \frac{2}{19} \times ( - \frac{7}{1} )$

We multiply to get

$\frac{ - 2 \times - 7}{19 \times 1}$

This simplifies to :

$\frac{14}{19}$

The correct answer is C

7 0

3 years ago

Read 2 more answers

a toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its heigh

sdas [7]

Answer:

t = 2.11 seconds

Step-by-step explanation:

A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$