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2 years ago

State the postulate or theorem you would use to prove each pair of triangles congruent.

Mathematics

1 answer:

quester [9]2 years ago

6 0

Answer:

c) ASA

Step-by-step explanation:

Angles at either end of the shared line segment are marked congruent, so the triangles can be shown to be congruent using the ASA postulate. (The shared segment is congruent to itself.)

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Solvethe equation 3(x-5)/2=2(x+5)/3

Nataliya [291]

3(x−5)/2=2(x+5)/3

Cross-multiply.

(3(x−5))/2=(2(x+5))/3

3(x−5)*(3)=2(x+5)*(2)

9x−45=4x+20

Subtract 4x from both sides.

9x−45−4x=4x+20−4x

5x−45=20

Add 45 to both sides.

5x−45+45=20+45

5x=65

Divide both sides by 5.

5x/5=65/5

x=13

6 0

3 years ago

Read 2 more answers

The diagram shows a cuboid.<br>4 cm<br>5 cm<br>9 cm<br>What is the surface area of the cuboid?

Salsk061 [2.6K]

Answer:

69cm ^2

Step-by-step explanation:

4 0

2 years ago

Point M is the midpoint of CD¯¯¯¯¯¯¯¯. C has coordinates of (-1, 1) and D has coordinates of (5,-5). What are the coordinates of

snow_lady [41]

Answer:

The answer is (2,-2)

7 0

2 years ago

John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart

Rina8888 [55]

<h2>Answer:</h2>

$Probability=\frac{1}{24}$

<h2>Step-by-step explanation:</h2>

As the question states,

John's brother has Galactosemia which states that his parents were both the carriers.

Therefore, the chances for the John to have the disease is = 2/3

Now,

Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

Now, one of those children married with a person.

So,

Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

Therefore,

<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

$Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}$

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)

<em><u>Hence, the probability for the first child to have Galactosemia is $\frac{1}{24}$ </u></em>

3 0

3 years ago

1. A pond has 85 frogs, 20% of the frogs are green the rest are orange, brown, and yellow. How many frogs are green?

alexandr402 [8]

Answer:

17 green frogs

Step-by-step explanation:

You need to multiply the decimal form of 20% by 85.

To find the decimal form of a percent you take the decimal (20.00) and move it twice to the left (20.00 ---> .2000)

Now, multiply by .2 by 85

(.2)(85) = 17

Therefore, 17 frogs in the pond of 85 frogs are green

<em>Hope this helps!!</em>

4 0

3 years ago