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nadya68 [22]
2 years ago
12

Can i get some help with these 2 problems

Mathematics
1 answer:
denis-greek [22]2 years ago
8 0
I hope this helps you

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What is an equation of the line that passes through the point (-4, -6) and is
const2013 [10]
Answer:y=2/3x-10/3
- 4/-6 is 4/6 so m is 4/6 then fill in the x and y variables in y=Mx+b solve for b then start plotting the equation of the line to check and you get your answer

I think this is the answer you might need to check again with more people
8 0
2 years ago
Read 2 more answers
This is probably easy to answer but I’m tired soo...
Flura [38]

Answer:

6:30

Step-by-step explanation:

20+16+6= 42

72-42=30

there are 30 rocks that are slate

8 0
3 years ago
Read 2 more answers
The table below represents the function f
alukav5142 [94]

Answer:

1

Step-by-step explanation:

I belive its one because 3^2  is 9 and 9+1 = 10

to make sure  it not just once 3^3 is 27 and 27+1=28

3 0
3 years ago
The factorization of x2 + 3x – 4 is modeled with algebra tiles. An algebra tile configuration. 2 tiles are in the Factor 1 spot:
arsen [322]

Answer:

The factors of x^2+3x-4 are (x-1)(x+4) ....

Step-by-step explanation:

We have to find the factors of x^2+3x-4

As we know that this is a quadratic equation.

So we have to find the roots first.

The roots are -1 and 4.

Now completing the quadratic formula using the roots we have :

x^2+4x-x-4

Make a pair of first two terms and last two terms:

(x^2+4x)-(x+4)

Now take out the common from each pair:

x(x+4)-1(x+4)

(x-1)(x+4)

Thus the factors of x^2+3x-4 are (x-1)(x+4) ....

5 0
3 years ago
Read 2 more answers
The cost per week of running a boarding house is partly constant and partly
patriot [66]

Answer:

the cost of running the boarding

house for 600 students is N61,000

Step-by-step explanation:

Let C represents cost

K1 represents first constant

K2 represents second constant

C= k1+k2n

3500 = k1 + 25 k2............. Eqn(1)

6000= k1 + 50 k2 .............. Eqn(2)

Subtract eqn(1) from eqn(2)

2500= 25k2

K2= 2500/25

K2= 100

To get k1 from eqn(1)

3500 = k1 + 25 k2

Substitute the value of k2

3500 = k1 + 25 (100)

3500= k1 +2500

K1= 3500- 2500

K1= 1000

The equation connecting them;

C= 1000+ 100n

The cost of running the boarding

house for 600 students is

n= 600

C= 1000+ 100(600)

C= 1000+60000

C= N 61,0000

6 0
2 years ago
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