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Art [367]
2 years ago
10

Solve the system by substitution

Mathematics
1 answer:
Tatiana [17]2 years ago
8 0

Answer:

(x,y) = (-36,6)

Step-by-step explanation:

-6y=x

x+y=-30

x=-6y,

so -6y+y=-30

-5y=-30

y=6

x=-36

Hope this helps pls hit the crown :D

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Tom plants 3 seeds.
Rasek [7]

Answer:

A. 64/125

B. 124/125.

Step-by-step explanation:

A).  As the events ( germinate or not germinate) are independent we multiply the probabilities.

Prob(All seeds germinate) = 4/5*4/5*4/5 =  64/125.

B). Probability of at least one germinating =  1 - probability that none germinate

Probability of  1 seed not germinating = 1 -45 = 1/5.

So Prob(at least one germinating)

= 1 - (1/5 * 1/5 * 1/5)

= 1 - 1/125

= 124/125.

5 0
3 years ago
X2 - 5x + 4<br> What is equivalent to this expression?
drek231 [11]

Answer

(x-1) (x-4)

Step-by-step explanation:

I did this not long ago in class so that's the answer

6 0
2 years ago
Read 2 more answers
Will do a brainly! 4/7=18/31.5
german

Answer:

True

Step-by-step explanation:

  1. 4 ÷ 7 = 0.5714
  2. 18 ÷ 31.5 = 0.5714
  3. Because they are the same result, the two fractions are equal. Therefore, the equation is true.

I hope this helps!

3 0
3 years ago
In Which Quadrant is this true
34kurt

Given:

\sin \theta

\tan \theta

To find:

The quadrant in which \theta lie.

Solution:

Quadrant concept:

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II, only \sin\theta and \csc\theta are positive.

In Quadrant III, only \tan\theta and \cot\theta are positive.

In Quadrant IV, only \cos\theta and \sec\theta are positive.

We have,

\sin \theta

\tan \theta

Here, \sin\theta is negative and \tan\theta is also negative. It is possible, if \theta lies in the Quadrant IV.

Therefore, the correct option is D.

5 0
2 years ago
Please answer this ASAP. The question is down below. Thank you!
Sonja [21]

Answer:

y = .5x^2 -2x -5

Step-by-step explanation:

Well we can start by seeing if the parabola is the same width by comparing it to its parent function ( y = x^2 )

In y = x^2 the 2nd lowest point is just up 1 and right 1 away from the vertex.

This is not true for our parabola.

So we can widen it by to the desidered width by making the x^2 into a .5x^2.

So far we’ve got y = .5x^2

Now the parabola y intercept is at -5.

So we can add a -5 into the equation making it.

y = .5x^2 - 5

Now for the x value.

So we can find the x value by seeing how far away the parabola is from from the y axis.

So the x value is -2x.

So the full equation is y = .5x^2 -2x -5

Look at the image below to compare.

7 0
3 years ago
Read 2 more answers
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