8th grade geometry HELP PLEASE THIS IS ALREADY LATE AND IS COMPLETELY DUE SOON

Please help me.Find the value of x in the isosceles triangle shown below

blagie [28]

There is an isosceles triangle.

The triangle is like two right triangles put together.

To find out the height of the triangle, we remove the other right triangle and we use the other triangle to find the height using the Pythagorean theorem.

Phytagoras Theorem :

b² = c² - a²

x² = 8² - (6÷2)²

x² = 64 - 3²

x² = 64 - 9

x² = 55

x = √55

So, the value of x is √55 (B)

Hope it helpful and useful :)

4 0

3 years ago

X+9^4=[4234+56/23] x94 = ? pelase show work

Lana71 [14]

Answer:

391,660.6

Step-by-step explanation:

First we need to do what is in the parentheses. 56/23 equals 2.4. 4234 + 2.4 is 4236.4... 4236.4 times 94 is 398,221.6... Now 9 to the fourth power is 6561. x + 6561 is 398,221.6 so x equals 391,660.6... I hope this helps and please make me the brainliest. Thanks!

3 0

3 years ago

Look at this expression

vladimir2022 [97]

Answer:

OD22264 is the correct answer for that

4 0

3 years ago

I don’t know how to do these problems so please help me.

Olin [163]

Answer:

36s^3

Step-by-step explanation:

3 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago