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olganol [36]
2 years ago
6

(01.01 LC) Evaluate the expression 3(7 + 4)2 - 14 - 7.

Mathematics
1 answer:
Flauer [41]2 years ago
6 0

Answer:

45

Step-by-step explanation:

3(11)2-14-7

33*2-14-7

66-14-7

52-7

45

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Help a gal out, A= *I have no idea* ?<br><br> Give me what A= and I will appreciate it so much!!!
sineoko [7]

Answer:

\frac{sin (c)}{AB}=\frac{sin(A)}{26}

sin(A)=\frac{26\times sin(13)}{6} =0.97

A=sin^{-1}(0.97)=\boxed{77^o}

6 0
2 years ago
Multiply: (x+4)(x^2+2x+3) <br> Simplify and write your answer in Standard Form.
Oxana [17]

Answer:

x^3+6x^2+11x+12

Step-by-step explanation:

(x+4)(x^2+2x+3)

=x(x^2+2x+3)+4(x^2+2x+3) [by multiplying with bith sides]

=x^3+2x^2+3x+4x^2+8x+12

=x^3+2x^2+4x^2+3x+8x+12

=x^3+6x^2+11x+12

(please mark me brainliest)

6 0
1 year ago
SAT scores probability 1​
fredd [130]

Answer:

  a.  81.5%

Step-by-step explanation:

The z-score for 400 is ...

  Z = (X -μ)/σ = (400 -500)/100 = -1

The z-score for 700 is ...

  Z = (700 -500)/100 = 2

The empirical rule tells you that 68% of the distribution is within ±1σ of the mean, and 95% is within ±2σ of the mean. Half of that first number is in the range Z = -1 to 0, and half that second number is in the range Z = 0 to +2. So, the probability you want is ...

  (1/2)(68%) + (1/2)(95%) = 81.5% . . . . matches choice A

6 0
3 years ago
Write a linear equation in slope-intercept form: 5y – 3x + 20 = 0 ​
Arturiano [62]

Slope intercept form is y = mx+ b

So in order to turn 5y-3x + 20 = 0 into Slope intercept form, all we need to do is solve for y!

Let's do it!

  • add 3x to the right side of the equation

<u>5y - 3x + 20 = 0</u>

<u>5y + 20 = 3x </u>

  • move 20 to the right side

<u>5y = 3x - 20 </u>

  • Divide both sides of the equation by 5

y = 3/5x - 4

This is our answer!

I hope I helped! Leave a comment if you have any questions or concerns

8 0
2 years ago
F(x)=X over x^3-2x^2+5x why will this have no zeros​
Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

\dfrac{x}{x^3-2x^2+5x}=0\iff x=0

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}

So, we can't even extend with continuity this function in such a way that f(0)=0

5 0
3 years ago
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