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2 years ago

Simplify in standard form 10a(-6a^2+2a-7)

Mathematics

1 answer:

oee [108]2 years ago

4 0

-60a^3+20a^2-70a=Answer

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hoa [83]

I hope this helps, If there are 5 cups of flour, then there will be 7 and 1/2 cups of sugar:)

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2 years ago

If 2a/4=b/4 and b≠0, what does 5b equal in terms of a?

dedylja [7]

Step-by-step explanation:

you must have made a typo here.

none of the answer options are correct for the given problem.

let me just show you what you told me, and how I can solve at least this :

2a/4 = b/4

this simply means (multiply both sides by 4) :

2a = b

so, 5b is then (multiplying both sides by 5) :

5×2a = 5b

10a = 5b

but again, "10a" is not among the offered answers.

so, I don't know if you made a mistake with the basic problem or with the answer options.

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1 year ago

What is a monthly expense you should budget for when buying a car?

AfilCa [17]

Answer:

insurance payments every month

8 0

2 years ago

Read 2 more answers

Please help asap!!!!!!!!!!!

mars1129 [50]

X = 59, y = 44

2x + 18 + x - 15 = 180
3x + 3 = 180
3x = 177
x = 59

x - 15 = y
59 - 15 = y
44 = y

6 0

2 years ago

Read 2 more answers

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

4 0

2 years ago