Ask question

Ask question

All categories

3 years ago

11

Please help me now plz I will mark you brainliest

1 answer:

cricket20 [7]3 years ago

8 0

Answer:

B.) x=5

Step-by-step explanation:

Use the Pythagorean theorem. Since it isosceles, both values of x are the same. Split 6 in half (because perpendicular bisector).Insert values:

$a^2+b^2=c^2\\3^2+4^2=x^2\\9+16=x^2\\25=x^2\\5=x$

X is equal to 5.

Done.

You might be interested in

The map shows the location of the airport and a warehouse in a city. Though not displayed on the map, there is also a factory 72

lisov135 [29]

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\sqrt{(-27-27)^2+(36+36)^2}\\\\ \sqrt{54^2+72^2} \\\\ \sqrt{8100} \\\\90\\\\\\90+54 =\boxed { 144}\\\\\sf The~ distance~ 90~ is~ from ~the~ warehouse~ to ~the ~airport\\\\The~ distance~ 54~ is~ from ~the~ airport~ to ~the~ factory\\\\So,~ the~ total~ distance~ is~ 144$

6 0

3 years ago

Read 2 more answers

Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides of th

Keith_Richards [23]

Answer:

8 cm and 9 cm

Step-by-step explanation:

The other two sides added together have to equal a larger number than 13, since 13 is the largest side.

5 + 8 = 13

6 + 7 = 13

7 + 2 = 9

8 + 9 = 17

7 0

3 years ago

Read 2 more answers

A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla

nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = $P(A)= \frac{13}{100}=0.13$

The probability that athlete is a basketball player = $P(B)= \frac{52}{100}=0.52$

The probability that athlete is both basket ball player and football player = $P(A\cap B) = \frac{9}{100}=0.09$

We have to find the probability that an athlete chosen is either a football player or a basketball player $P(A\cup B)$ .

Now we know that;

$P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%$

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0

3 years ago

If arc AB = 107° and arc BC = 122°, find the value of x.

Aneli [31]

Answer:

the value of x is acb 133°

4 0

2 years ago

Use the formula to calculate the present value of each payment, assuming an interest rate of 3%. The first several payments have

Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

234

4 0

3 years ago