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sasho [114]
3 years ago
6

Which ordered pair is the solution to the system of linear equations -5x+y=26 and 2x-7y=16

Mathematics
1 answer:
Yakvenalex [24]3 years ago
3 0
The answer is D. (-6,-4)
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Which two fractions name point R on the number line?
3241004551 [841]

Answer:

The answer is D.) 6/8 and 3/4

Step-by-step explanation:

If you count how many lines there are you get 8 lines so we know the denominator has to be eight and since the point is on the 6th line the first one is 6/8 and since you can simplify it you get 3/4 which gives you D.)

Hope I helped

A brainliest is always appreciated

4 0
2 years ago
Write an equation that you can use to solve for x.<br><br> Enter your answer in the box.
mario62 [17]

Answer:

40 degrees

100-60=40

Step-by-step explanation:

The angle x is in another angle that looks like 100 and 100 minus 60 is 40.

Hope this helps and if it is correct.

4 0
2 years ago
Read 2 more answers
Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the
Virty [35]

Answer:

V1 = 60 km/h

V2 = 40 Km/h

Step-by-step explanation:

The speed of an object is defined as

Speed =  distance / time

Let

V1 be the speed of the faster car

V2 be the speed of the other car

t1 the time it took for the first car to arrive

t2 the time it took for the second car to arrive

d1 the distance traveled by first car

d2 the distance traveled by second car

We know thanks to the problem that

V1 = V2 + 20 Km/h

t1 = t2 - 1 hour

d1 = d2 = 120 Km

d1 = V1 * t1

d2 = V2* t2

V1 * t1 = V2* t2

V1* t1 = (V1 -20)*(t1 +1)

The system of equations

(V1 -20)*(t1 +1) = 120

V1 * t1 = 120

120 + (120/t1) -20*t1 = 140

(120/t1) -20*t1  = 20

Which gives,

t1 = 2

This means

V1 = 60 km/h

V2 = V1 - 20 Km/h =  40 Km/h

6 0
3 years ago
See attached picture, is there a way to enter equations or algebraic symbols?
coldgirl [10]

SOLUTIONS

Solve a given equations or algebraic symbols?

\begin{gathered} f(x)=x^2+2x \\ g(x)=1-x^2 \end{gathered}

(A)

\begin{gathered} (f+g)(x)=(x^2+2x)+(1-x^2) \\ collect\text{ like terms} \\ x^2-x^2+2x+1 \\ (f+g)(x^)=2x+1 \end{gathered}

(B)

\begin{gathered} (f-g)(x)=(x^2+2x)-(1-x^2) \\ =x^2+2x-1+x^2 \\ =x^2+x^2+2x-1 \\ (f-g)(x)=2x^2+2x-1 \end{gathered}

(C)

\begin{gathered} fg(x)=(x^2+2x)(1-x^2) \\ =x^2-x^4+2x-2x^3 \\ fg(x)=-x^4-2x^3+x^2+2x \end{gathered}

(D)

\begin{gathered} \frac{f}{g}(x)=(x^2+2x)\div(1-x^2) \\ =\frac{x^2+2x}{1-x^2} \end{gathered}

5 0
1 year ago
Please someone help me with this please
Elodia [21]
Help u with what I need to know what u need help on
3 0
3 years ago
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