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Lubov Fominskaja [6]
2 years ago
8

Brainliest to whoever answers first!!!

Mathematics
2 answers:
Verizon [17]2 years ago
8 0

Answer:

11x^2+4x-6

Step-by-step explanation:

4x^2+9x-7+7x^2-5x+1=11x^2+4x-6

You combine the like terms.

alexgriva [62]2 years ago
4 0

Add/Subtract Like terms

Answer:

11x^2+4x-6

Hope this helped :)

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Find the area Hint:Triangle and semi-circle
sergeinik [125]

Answer:

The triangle formula is always base times height divided by 2 I forgot the circle formula

Step-by-step explanation:

8 0
2 years ago
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If (x + 2) is a factor of x3 − 6x2 + kx + 10, k =
ch4aika [34]

Answer:

The value of k is -11

Step-by-step explanation:

If (x+2) is a factor of x3 − 6x2 + kx + 10:

Then,

f(x)=x3 − 6x2 + kx + 10

f(-2)=0

f(-2)=(-2)³-6(-2)²+k(-2)+10=0

f(-2)= -8-6(4)-2k+10=0

f(-2)= -8-24-2k+10=0

Solve the like terms:

f(-2)=-2k-22=0

f(-2)=-2k=0+22

-2k=22

k=22/-2

k=-11

Hence the value of k is -11....

5 0
2 years ago
Whats x squared -4x+4
Dafna11 [192]

Answer:

Step-by-step explanation:

ajeroeriwrawr

7 0
2 years ago
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Three hundred twenty four million in numbers
LenaWriter [7]

24,000,300

Step-by-step explanation:

you mean twenty-four million and three hundred right? or Twenty million and three hundred thousand?

because that's not possible.

5 0
3 years ago
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If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
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