Answers for number 8 and 11 ?

Two corresponding sides of similar triangles are 2 cm and 5 cm. What is the area of the second triangle if the area of the first

Stella [2.4K]

Answer:

(1) Area of second triangle= 50 cm^2.

(2) Area of ΔABC=98 cm^2.

and Area of ΔDFG=175 cm^2.

Step-by-step explanation:

<em>" For two similar triangles the ratio of sides is equal to the ratio of square of their areas".</em>

i.e. if a,b are the corresponding sides of two similar triangles and let A and B denote the area of two triangles then we have the relation as:

$\dfrac{a^2}{b^2}=\dfrac{A}{B}$

(1)

for the first question:

we have a=2, b=5.

A=8 cm^2.

Hence,

$\dfrac{2^2}{5^2}=\dfrac{8}{B}\\\\\dfrac{4}{25}=\dfrac{8}{B}$

B=50 cm^2.

Hence, the area of second triangle is 50 cm^2.

(2)

In second option we have:

a=6 and b=5.

A-B=77 cm^2.

A=77+B

$\dfrac{6^2}{5^2}=\dfrac{A}{B}\\ \\\dfrac{36}{25}=\dfrac{77+B}{B}\\ \\36B=25\times77+25B\\\\36B-25B=25\times77\\\\11B=25\times77\\\\B=25\times7\\\\B=175$

Hence area of second triangle i.e. ΔDFG is 175 cm^2.

and area of first triangle i.e. ΔABC=175-77=98 cm^2.

8 0

3 years ago

The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of

SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20.8, \sigma = 4.5$

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 20.8}{4.5}$

$Z = 0.04$

$Z = 0.04$ has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23 - 20.8}{4.5}$

$Z = 0.71$

$Z = 0.71$ has a pvalue of 0.7611

X = 18

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{18 - 20.8}{4.5}$

$Z = -0.62$

$Z = -0.62$ has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0

3 years ago

Hurry will give brainiest.

Iteru [2.4K]

Answer:

2. z=8

3.2x=16

5.distributive property

7.Addition property of equality

i'm not 100%ly sure so if it's wrong sorry and pls don't report mine i'm just trying to help not get my answer deleted:)

8 0

3 years ago

The area of sector A08 is 48x and m<AOB = 270º. Find the radius of circle O.

Softa [21]

Answer:

Step-by-step explanation:

The first one gives us everything we need except the radius, which is easy enough to solve for if you're careful with your algebra. The area of a sector of a circle is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ is the measure of the central angle of the circle. For us, that fills in as follows:

$48\pi=\frac{270}{360}*\pi r^2$ and manipulate it as follows:

$r^2=\frac{(360)(48\pi)}{270\pi}$ the π's cancel out, leaving us with simple multiplication and division to get

r = 8. Now for the next one, which is a bit more involved.

In order to find the area of the shaded part, we need to find the area of the right triangle there and subtract it from the area of the sector of the circle. First the area of the sector, which is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ again is the measure of the central angle of the circle, 90°:

$A_s=\frac{90}{360}(3.1415)(10)^2$ which simplifies a bit to

$A_s=\frac{1}{4}(3.1415)(100)$ , giving us an area of

$A_s=78.5375m^2$ . Now onto the area of the triangle.

Since this triangle is inscribed in the circle and the circle's radius is 10, tha also gives us both the height and the base measures of the triangle. The area then is:

$A_t=\frac{1}{2}(10)(10)$ which is

$A_t=50m^2$

Subtract that from the area of the sector to get that the shaded area is 28.5 square meters, choice A.

4 0

4 years ago

3.<br> 27x3+8<br> Type(s): _________

Novosadov [1.4K]

27 times 3= 81 then add 9 which then equals 89

6 0

4 years ago