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2 years ago

Find the length of QT if Q is between M and T, if MT=18.3 and MQ=7.8

Mathematics

1 answer:

creativ13 [48]2 years ago

8 0

Answer: QT= 10.5

Step-by-step explanation:

18.3-7.8

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Vedmedyk [2.9K]

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$f(x) + g(x) = 5^x + 2x + 3x -6 = 5^x + 5x - 6$

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Jerry works for 7 hours at $10.50 per hour. How much does she earn

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Answer: Jerry earns $73.50

Step-by-step explanation:

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10.50n

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3 years ago

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A statue is mounted on top of a 21 foot hill. From the base of the hill to where you are standing is 57feet and the statue subte

AleksandrR [38]

Please find the attached diagram for a better understanding of the question.

As we can see from the diagram,

RQ = 21 feet = height of the hill

PQ = 57 feet = Distance between you and the base of the hill

SR= h=height of the statue

$\angle SPR$ =Angle subtended by the statue to where you are standing.

$\angle x=\angle RPQ$ = which is unknown.

Let us begin solving now. The first step is to find the angle $\angle x$ which can be found by using the following trigonometric ratio in $\Delta PQR$ :

$tan(x)=\frac{RQ}{PQ} =\frac{21}{57}$

Which gives $\angle x$ to be:

$\angle x=tan^{-1}(\frac{21}{57})\approx20.22^{0}$

Now, we know that $\angle x$ and $\angle SPR$ can be added to give us the complete angle $\angle SPQ$ in the right triangle $\Delta SPQ$ .

We can again use the tan trigonometric ratio in $\Delta SPQ$ to solve for the height of the statue, h.

This can be done as:

$tan(\angle SPQ)=\frac{SQ}{PQ}$

$tan(7.1^0+20.22^0)=\frac{SR+RQ}{PQ}$

$tan(27.32^0)=\frac{h+21}{57}$

$\therefore h+21=57tan(27.32^0)$

$h\approx8.45 ft$

Thus, the height of the statue is approximately, 8.45 feet.

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2 years ago

3 4/7 minus 2 1/5 by rounding to the nearest half

mylen [45]

Answer:

$1 \frac{13}{35}$

Step-by-step explanation:

$3 \frac{4}{7} - 2 \frac{1}{5}$

you can solve for the fractions first, but you have to give them a common denominator

$\frac{20}{35}- \frac{7}{35}$

$= \frac{13}{35}$ which is in simplest form

then solve for the whole numbers

3-2=1

therefore, the answer is $1 \frac{13}{35}$

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2 years ago