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3 years ago

What is (-9)(-4)(-4)(8)

Mathematics

2 answers:

grandymaker [24]3 years ago

5 0

Answer:

-1152

Step-by-step explanation:

First: (-9)(-4)=36

Then: 36(-4)=-144

Lastly: -144(8)= -1152

Because It's multiplication we can separate the problem and break it down. By doing this is makes it a lot easier to understand and see the problem step-by-step.

MatroZZZ [7]3 years ago

4 0

Answer:

-1,152

Step-by-step explanation:

(-9)(-4)(-4)(8)

= -1152

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Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

EleoNora [17]

Answer:

For the perimeters, x must be equal to 2.

For the areas, it is either undefined, or something.

Step-by-step explanation:

You can first find the perimeters for both sides.

For the left shape, we add the two sides of 6 and x + 4 to get x + 10.

Then we multiply x + 10 by 2 because there are 4 sides, and we only got 2 sides.

The perimeter of the first shape is 2x + 20.

The second shape can be solved by doing the same thing by adding 2 and 3x + 4 to get 3x + 6.

3x + 6 times 2 is 6x + 12.

The second perimeter is 6x + 12.

If both sides are supposed to be equal, then we can write these two expressions we solved for like:

6x + 12 = 2x + 20.

Subtraction property of equality

6x + 12 - 12 = 2x + 20 - 12

Simplify

6x = 2x + 8

Again

6x - 2x = 2x - 2x + 8

Simplify

4x = 8

Division property of equality

4/4x = 8/4

Simplify

x = 2

So if x = 2, the perimeters will be the same.

You can confirm this by plugging it back into either equation.

For the areas, we just multiply the length and width for both shapes, so we get

6(x+4) = 2(3x+4)

Since they are supposed to be equal.

We simplify and get

6x + 24 = 6x + 8

We know this is false and is not possible, since we can remove the 6x because it is on both sides.

We also know that 24 is not equal to 8 (who thought!)

24 ≠ 8

So it is undefined or whatever you call it.

3 0

3 years ago

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

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3 years ago

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telo118 [61]

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mina [271]

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Montano1993 [528]

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