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krek1111 [17]
3 years ago
15

Adam currently runs about 20 miles per week, and he wants to increase his weekly mileage by 30%. Howmany miles will Adam run per

week ?
Mathematics
2 answers:
Vadim26 [7]3 years ago
8 0

Answer:

50 %

Step-by-step explanation:


andrew-mc [135]3 years ago
6 0
Adam will have to run 26 miles per week. 30% of 20 is 6. (6+20=26)
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2. A bag contains 3 red, 4 white, and 5 blue marbles. Jason begins removing marbles from the bag at
galben [10]

Answer:

c

the total number of marbles are 12 and there are 3 colours so you divide 12 by 3

6 0
3 years ago
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PLEASE HELP!! NO LINKS!! GIVING 20 POINTS AWAY!!
Kruka [31]

Answer:

m∠N = 32°

NQ = 106°

When finding inscribed angles like ∠N with the intercepted arc, the equation is ∠N=1/2MP. (Inscribed angles are always half the degree of the arc length.) Plug in the corresponding value to get ∠N=1/2(64) to get 32°. When finding the angle of the intercepted arc with inscribed angles like NQ, the equation is NQ=2(∠P). Plug in the corresponding value to get 2(53) to get 106°.

5 0
3 years ago
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Which is the solution to the equation solved for p? (q + p)q = (s + r)s
Brums [2.3K]

Answer:

Option B

Step-by-step explanation:

(q + p)q = (s + r)s

Lets open the brackets of L.H.S.

=> q^2 + pq = (s+ r)s

Lets substract q² from both the sides.

=> q^2 + pq - q^2 = (s + r)s - q^2

=> pq = (s + r)s - q^2

Lets divide both the sides by q.

=> \frac{pq}{q} = \frac{(s + r)s - q^2}{q}

=> p = \frac{(s + r)s }{q} - \frac{q^2}{q} = \frac{(s + r)s}{q} - q

7 0
2 years ago
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
frozen [14]

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

(x,y)\to (y,-x)

Using this rule, we get

A(-3,6)\to A'(6,3)

B(2,9)\to B'(9,-2)

C(1,1)\to C'(1,-1)

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

7 0
3 years ago
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Aliun [14]

Answer:

is something supposed to be there

Step-by-step explanation:

4 0
2 years ago
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