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2 years ago

Based on the angle measures what type of triangle is ABD

Mathematics

1 answer:

vodomira [7]2 years ago

8 0

Isosceles triangle because there are 2 angles that's the same.

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Which of the following figures has a length, width, and a height

astra-53 [7]

C. Cube
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7 0

2 years ago

Read 2 more answers

The areas of the two watch faces have a ratio of 16:25 What is the ratio of the radius of the smaller watch face to the radius o

Leona [35]

Answer:

The ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

Step-by-step explanation:

Let the Area of smaller watch face be $A_1$

Also Let the Area of Larger watch face be $A_2$

Also Let the radius of smaller watch face be $r_1$

Also Let the radius of Larger watch face be $r_2$

Now given:

$\frac{A_1}{A_2} =\frac{16}{25}$

We need to find the ratio of the radius of the smaller watch face to the radius of the larger watch face.

Solution:

Since the watch face is in circular form.

Then we can say that;

Area of the circle is equal 'π' times square of the radius 'r'.

framing in equation form we get;

$A_1 = \pi {r_1}^2$

$A_2 = \pi {r_2}^2$

So we get;

$\frac{A_1}{A_2}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Substituting the value we get;

$\frac{16}{25}= \frac{\pi {r_1}^2}{\pi {r_2}^2}$

Now 'π' from numerator and denominator gets cancelled.

$\frac{16}{25}= \frac{{r_1}^2}{{r_2}^2}$

Now Taking square roots on both side we get;

$\sqrt{\frac{16}{25}}= \sqrt{\frac{{r_1}^2}{{r_2}^2}}\\\\\frac{4}{5}= \frac{r_1}{r_2}$

Hence the ratio of the radius of the smaller watch face to the radius of the larger watch face is 4:5.

7 0

3 years ago

A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in

nordsb [41]

Answer:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

$h(t) = -6.3t^2 +53 t+24$

In order to maximize this function we need to take the first derivate respect the time and we have:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

8 0

3 years ago

Find the number that makes the ratio equivalent to 1:6.<br> 9:__

klio [65]

Okay, what do we know?

1*6=6, right?

Well, then 9*6=? (you should know your math facts by now)

3 0

2 years ago

Plz help and explain

Firdavs [7]

Id say the first one because the points just touch down

8 0

3 years ago