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BaLLatris [955]
4 years ago
6

Help me this question... is easier one

Mathematics
1 answer:
lara [203]4 years ago
8 0

a)x + 3 = 0 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x =  - 3

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Plz help.........................
Alex17521 [72]
The answer would be d 
7 0
4 years ago
<img src="https://tex.z-dn.net/?f=HELP%20%5C%3A%20%20ME%20%5C%3A%20%20PLEASE%20" id="TexFormula1" title="HELP \: ME \: PLEASE
Olin [163]

Answer:

22°

Step-by-step explanation:

as we know from the angle sum property of the triangle- the sum of all angles of a triangle is 180°

So now in (d)

79°+79°+x=180°

158°+x=180°

x=180°-158° (transversal property)

x=22°

Hope it helps

8 0
3 years ago
Read 2 more answers
Evaluate the expression: 15c4<br> A. 8,190<br> B. 32,760<br> C. 1,365<br> D. 5,460
WARRIOR [948]

Answer:

C. 1365

Step-by-step explanation:

The expression nCr can be written with the simple equation:

n! / ((n-r)! * r!)

So to evaluate 15C4:

15! / ((15-4)! * 4!)

= 1365

So 15C4 = 1365

Cheers.

6 0
4 years ago
Read 2 more answers
Each grade level at Hooperville Schools has 298 students.
madreJ [45]

Answer:

(a) 3,874 students.

(b) 46,488 students.

Step-by-step explanation:

We have been given the each grade level at Hooperville Schools has 298 students.

(a) To find the number of students that attend Hooperville Schools, we will multiply the number of grades by 298.

\text{Students that attend Hooperville Schools}=13\times 298

\text{Students that attend Hooperville Schools}=3,874

Therefore, 3,874 students attend Hooperville Schools.

(b) We have been given that nearby district, Willington, is much larger. They have 12 times as many students.

To find the number of students that attend schools in Willington, we will multiply number of students at Hooperville schools by 12.

\text{Students that attend Willington Schools}=12\times 3,874

\text{Students that attend Willington Schools}=46,488

Therefore, 46,488 students attend schools in Willington.

4 0
3 years ago
Suppose you have 20 coins that total $4.some coins are nickles and sime are quarters .which of the following pairs of equations
Monica [59]
So 1 nickle=5 cents
1 quarter=25 sents

q=number of quarters
n=number of nickles so

total is
q+n=20
5n+25q=400 (since 100 cents=1 dollar and 4 dollares=400 cents)

SO THE 2 EQUATIONS WOULD BE
n+q=20
5n+25q=400

so q+n=20
subtract q from both sides
n=20-q
subsitute 20-q for n in the second equation
5(20-q)+25q=400
distribute
100-5q+25q=400
add like terms
100+20q=400
subtract 100 from both sides
20q=300
divide both sides by 20
q=15

thhere are 15 quarters

subsitute 15 for q in first equation

15+n=20
subtract 15 from both sides
n=5

5 nickles
15 quarters
4 0
4 years ago
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