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VladimirAG [237]

2 years ago

15

Read the word problem below. Chan rows at a rate of eight miles per hour in still water. On Wednesday, it takes him three hours

to row upstream from his house to the park. He rows back home, and it takes him two hours. What is the speed of the current? Which of the following is a clue that would help solve this word problem? The rate of Chan’s rowing in still water is 8 mph. Chan rows from the house to the park on Wednesday. Chan’s house is next to a river. The entire trip takes five hours.

1 answer:

Romashka-Z-Leto [24]2 years ago

3 0

Answer:

its A

Step-by-step explanation:

i literally just took the test

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En una cancha se va a pintar la circunferencia central que tiene un diámetro de 5 m .¿cual es la longitud de la circunferencia?

Alona [7]

Answer:

The length of the circumference is $5\pi\ m$ or $15.70\ m$

Step-by-step explanation:

<u><em>The question in English is</em></u>

On a court, the central circumference will be painted, which has a diameter of 5 m. What is the length of the circumference?

we know that

The circumference of a circle is given by the formula

$C=\pi D$

we have

$D=5\ m$

substitute

$C=\pi (5)\\C=5\pi\ m$

This is the exact value

assume

$\pi=3.24$

$C=5(3.14)=15.70\ m$

4 0

1 year ago

Please answer all of them need this

VikaD [51]

First Question

For a better understanding of the solution provided here please find the first attached file which has the diagram of the the isosceles trapezoid.

We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

As can be seen in $\Delta BCQ$ , we can easily find the values of CQ and BQ.

Since, $Sin(75^0)=\frac{CQ}{8}$

$\therefore CQ=8\times Sin(75^0)\approx 7.73$ ft

In a similar manner we can find BQ as:

$Cos(75^0)=\frac{BQ}{8}$

$BQ\approx2.07$ ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

$PQ=22-(AP+QB)=22-(2.07+2.07)=17.86$

Let us now consider $\Delta AQC$

We can apply the Pythagorean Theorem here to find the length of the diagonal AC which is the hypotenuse of $\Delta AQC$ .

$AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38$ feet.

Thus, out of the given options, Option B is the closest and hence is the answer.

Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

Area= $\frac{1}{2}$ (First Side)(Second Side)(Sine of the angle between the two sides)

Thus, from the given data,

Area= $\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845$ $m^2$

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the $\Delta ABC$ given to us.

Thus, from the triangle we will have:

$\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}$

$\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}$

$\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}$

This gives $a$ to be:

$a\approx28.44$

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

$AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}$

$\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8$ miles.

Thus, Option D is the answer.

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Colt1911 [192]

Answer:

Let x be the value of the house before increase

Since it's increases add 7% to the overall 100%

107/100x = 749000

x = 700000

The value of the house before increase is £700000.

Hope this helps.

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