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olasank [31]
2 years ago
13

This is hard for me can anyone help me? 9

Mathematics
1 answer:
Kazeer [188]2 years ago
5 0

Answer: -12.8x-20y+8.4

Step-by-step explanation: I multiplied the numbers in parentheses by 4.

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Please help,<br> True or False: <br><br> Will mark brainliest for the first person who answers
slamgirl [31]

Answer:

4.true

5.true

6.true

7.true

8.false

9.false

10. true

Step-by-step explanation:

I think, I'm sorry if some are wrong

3 0
3 years ago
HELP!!!!! Someone answer ASAP PLEASE!!!!
Andreas93 [3]

Answer:

thay are independent

Step-by-step explanation:

I only know that they are independent but Don't really know how to explain and do step by step.

7 0
3 years ago
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The graphs of g(x) = x³- ax² + 6 and h(x) = 2x² + bx + 3 touch when x = 1. Therefore, the tangent to
Olegator [25]

Answer:

(1,\, 10).

Step-by-step explanation:

Differentiate each function to find an expression for its gradient (slope of the tangent line) with respect to x. Make use of the power rule to find the following:

g^\prime(x) = 3\, x^2 - 2\, a\, x.

h^\prime(x) = 2\, (2\, x) + b = 4\, x  + b.

The question states that the graphs of g(x) and h(x) touch at x = 1, such that g^\prime(1) = h^\prime(1). Therefore:

3 - 2\, a = 4 + b.

On the other hand, since the graph of g(x) and h(x) coincide at x = 1, g(1) = h(1) (otherwise, the two graphs would not even touch at that point.) Therefore:

1 - a + 6 = 2 + b + 3.

Solve this system of two equations for a and b:

\begin{aligned}& a + b = 2 \\ & 2\, a + b = -1\end{aligned}.

Therefore, a = -3 whereas b = 5.

Substitute these two values back into the expression for g(x) and h(x):

g(x) = x^3 + 3\, x^2 + 6.

h(x) = 2\, x^2 + 5\, x + 3.

Evaluate either expression at x = 1 to find the y-coordinate of the intersection. For example, g(1) = 1 + 3 + 6 = 10. Similarly, h(1) = 2 + 5 + 3 = 10.

Therefore, the intersection of these two graphs would be at (1,\, 10).

6 0
2 years ago
Hey guys- need help in this one
oksano4ka [1.4K]

Answer:

c

Step-by-step explanation:

x=5+i\sqrt{7} , y = 5-i\sqrt{7}\\ x^2-2xy+y^2=(x-y)^2\\ (x-y)^2=(2i\sqrt{7} )^2\\4.i^2.7\\-28

8 0
2 years ago
Read 2 more answers
What is the length of the hypotenuse of the right triangle if one side is 16in and one is 7in?
Ronch [10]

Answer:

hypotenuse = 17.46

Step-by-step explanation:

The Pythagorean Theorem is a^2+b^2=c^2.

Since we are finding the hypotenuse, we must plug in 16 and 7 since they represent the lengths of both side.

16^2+7^2=c^2

256+49=c^2

305=c^2

which means we have to find the square root of c^2 AND 305, round it to the nearest hundredth, which means that;

c=17.46

7 0
3 years ago
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