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Naddik [55]
2 years ago
7

What is the value of 2(-8) (- 5) + 12

Mathematics
1 answer:
Vlad1618 [11]2 years ago
5 0
-68
That’s the answer
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I need help ASAP!!!!!
Alla [95]

 If we know that two sets of corresponding angles and the corresponding included sides are congruent in two triangles, what can we say about the triangles?


3 0
3 years ago
Which expression has a positive value?
Black_prince [1.1K]

Answer:

3 (negative 64 divided by 8) + 25 = 1

Step-by-step explanation:

Negative 4 + (negative 5) (negative 6) divided by (negative 3) = -4 + (-5)(-6)/(-3) = -4 + 30/-3 = -4 - 10 = -14

8 Left-bracket 10 divided by (2) (negative 2) Right-bracket = 8(10/((2)(-2)) = -20

3 (negative 64 divided by 8) + 25 = 3(-64/8) + 25 = 3(-8) + 25 = -24 + 25 = 1

Negative 2 (negative 5) (negative 3) divided by 10 = (-2)(-5)(-3)/10 = -3

The only positive value here is 1, which is a result of 3 (negative 64 divided by 8) + 25

6 0
3 years ago
Read 2 more answers
4-0,9+(-0,11)-(-1,3) PLEASE HELP
eduard

Answer:

-20

Step-by-step explanation:

PEMDAS

4 0
2 years ago
PLEASE HELP ASAP!! an onion soup recipe calls for 3 2/3 cups of chopped onions. katrina has already chopped 1 1/3 cups of onions
ivanzaharov [21]

Answer:

x+1\frac{1}{3}=3\frac{2}{3}

Step-by-step explanation:

Let the number of cups of onions Katrina still needs to chop = x

She has already chopped the onions = 1\frac{1}{3} cups ≈ \frac{4}{3} cups

So total amount of chopped onions = (x+\frac{4}{3}) cups

Since, onion soup recipe calls for the chopped onions = 3\frac{2}{3} cups ≈ \frac{11}{3} cups

Therefore, equation which shows the number of cups of onions to be chopped more will be,

x+1\frac{1}{3}=3\frac{2}{3}

We further solve this equation for the value of x,

x = \frac{11}{3}-\frac{4}{3}

x = \frac{(11-4)}{3}

  = \frac{7}{3} cups

  ≈ 2\frac{1}{3} cups

5 0
3 years ago
F(x) = x^2 + 1 is limited to (0, 1, 2, 3), what is the maximum love you all of the range?
wlad13 [49]

Answer:

2

Step-by-step explanation:

f(x)=x^2+1

now,

f(1)=1^2+1

=1+1

=2

4 0
3 years ago
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