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soldier1979 [14.2K]
3 years ago
11

Will mark brainliest!! please help me

Mathematics
1 answer:
lions [1.4K]3 years ago
7 0
The correct answer is A !
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Will i pass eighth grade with a d in math. give me a genuine answer too. not just to get points.
myrzilka [38]

yes you can

Step-by-step explanation:

D is a passing grade! ;)

3 0
2 years ago
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Proportions in Triangles
alukav5142 [94]

Answer:

x =35.

Step-by-step explanation:

x / 20 = 14 / 8

Cross multiply:

8x = 20*14

8x = 280

x = 35.

3 0
3 years ago
Pls help with #14 and #15!!!Asap 15 points
inna [77]

Answer:

14. -g + 10, 15. 12k + 3

Step-by-step explanation:

I'm assuming you want to simplify the equation. For 14, we distribute -2. This gets us -2g + 10 + g. We can simplify that to -g + 10. We can't simplify that, so our answer is complete. For 15, we combine 8k and 4k to get 12k + 3. That can't be simplified, so we are done.

8 0
3 years ago
PLEEEEASE PLAZE HELP ME WITH THIS! ITS A REALLY LATE ASSIGNMENT I FORGOT ABOUT AND ITS DUE TODAY AT 12:00! I WILL MARK AS BRINLI
Drupady [299]
1,800 is your answer because 1,200/72=16.6666667 and 1,800/108=16.6666667.


Hoped I helped!
7 0
3 years ago
1. ) Consider the function f(x)=5−7x2,−5≤x≤1
Marat540 [252]
1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.

Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170


Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2

2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

f'(c)= \frac{f(b)-f(a)}{b-a}

The differentiated form would be f'(x) = -2x. Then,

-2c =  \frac{(4- 0^{2} )-(4- (-1)^{2}) }{0--1}=1

c=- \frac{1}{2}

Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
3 0
3 years ago
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