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kaheart [24]
2 years ago
6

Choose the congruence theorem that you would use to prove the triangles congruent.

Mathematics
1 answer:
LekaFEV [45]2 years ago
7 0
The correct answer is SSS
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Can someone please help me on 20 and 21
Feliz [49]
For number 20 x = 4
3 0
3 years ago
Find the arc length of the partial circle.
nalin [4]

Answer:

1/2 pi

Step-by-step explanation:

This is 1/4 of a circle

Find the  circumference and multiply by 1/4

1/4 C

1/4 ( 2*pi*r)  where r is the radius

The radius is 1

1/2 pi*1

1/2 pi

7 0
3 years ago
Read 2 more answers
I'm having trouble with algebraic fractions . 3/2n-8/3=-29/12
nadezda [96]

what we can do is hmmmm get the LCD of all fractions, and multiply both sides by that LCD, let's see, we have denominators of 2, 3 and 12, so we can use the LCD of 12, and multiply both sides by 12 to do away with the denominators.

\bf \cfrac{3}{2}n-\cfrac{8}{3}=-\cfrac{29}{12}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{3}{2}n-\cfrac{8}{3} \right)=12\left( -\cfrac{29}{12} \right)}\implies (6)3n-(4)8=(1)-29 \\\\\\ 18n-32=-29\implies 18n=3\implies n=\cfrac{3}{18}\implies n=\cfrac{1}{6}

5 0
3 years ago
What is the circumference of a circle with diameter 24 m
marshall27 [118]

Answer:

24\pi

Step-by-step explanation:

4 0
3 years ago
A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm2/sec. Find the rat
lesya [120]

Answer:

The value of rate of which the base is changing \frac{dB}{dt} = - 3 \frac{cm}{s}

Step-by-step explanation:

Area = 20 cm^{2}

height = 4 cm

\frac{dH}{dt} = 2 \frac{cm}{sec}

\frac{dA}{dt} =  2 \ \frac{cm^{2} }{sec}

we know that area of the triangle is given by

A= \frac{1}{2} B H

B = \frac{2A}{H}

B = \frac{2 (20)}{4}

B = 10 cm

Rate of change of area is given by

\frac{dA}{dT} = \frac{1}{2} [B\frac{dH}{dt} + H \frac{dB}{dt} ]

4 =  0.5 [10 × 2 + 4 \frac{dB}{dt} ]

\frac{dB}{dt} = - 3 \frac{cm}{s}

This is the value of rate of which the base is changing.

7 0
3 years ago
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