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3 years ago

6

Compare a perpendicular bisector and an altitude of a triangle

1 answer:

lara31 [8.8K]3 years ago

3 0

Perpendicular bisectors are segment that passes through the mid point of a triangle, while altitude is the perpendicular distance from the base to the opposite vertex.

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The director of an athletic club wants

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Step-by-step explanation:

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3 years ago

shawn drew a rectangle that was 2 units wide and 6 units long..draw a different rectangle that has the same perimeter but a diff

defon

2(2)+2(6)= 16, so the perimeter must be 16. Let’s make one of the side lengths l = 3, so 2(3)+2(w)= 16, or w = 5. You can check by multiplying the dimensions that the areas are different.

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3 years ago

Jen is 7 cm shorter than twice Ann's height. together they are 146 cm. How tall is Jen?

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136 in cm I pretty sure i’m correct

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4 years ago

Read 2 more answers

To the nearest penny, what is the value of x?

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Answer: $43.20

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3 years ago

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Hector had 30 minutes to do a three-problem quiz. He spent 11 3 4 minutes on question A and 6 1 2 minutes on question B. How muc

stich3 [128]

Answer:

$11 \frac{3}{4}$ min

Step-by-step explanation:

In order to solve this problem, we must subtract the times he has already used from the time he has available, so we get:

$30-11 \frac{3}{4} - 6 \frac{1}{2}$

we can start solving this subtraction by turning all of the numbers into improper fractions so we get:

$\frac{30}{1}-\frac{11*4+3}{4}- \frac{6*2+1}{2}$

$\frac{30}{1}-\frac{47}{4}- \frac{13}{2}$

next, we can find a least common denominator. In this case it will be 4, so we need to turn all denominators into a 4, so we get:

$\frac{30*4}{1*4}-\frac{47}{4}- \frac{13*2}{2*2}$

$\frac{120}{4}-\frac{47}{4}- \frac{26}{4}$

and now we can subtract the numerators and copy the denominators so we get:

$\frac{120-47-26}{4}$

$\frac{47}{4}$

which can now be turned into a mixed number by dividing 47/4 which yields 11 with a remainder of 3, so the mixed number is:

$11 \frac{3}{4}$ min

and this is the amount of time he has available to solve the las problem.

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4 years ago