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3 years ago

What is the value of (10 x 0.5) squared + 4 cubed?

Mathematics

2 answers:

Salsk061 [2.6K]3 years ago

8 0

(10*0.5)= 5

5^2 = 25

25+4^3

25+64 = 89

kotegsom [21]3 years ago

4 0

First set up the equation:
$(10*0.5)^{2}$ + $4^{3}$

Now solve the ones in the parenthesis:
$5^{2}$ + $4^{3}$

Then do the exponents:
25+64

The answer will be: 89

Hope this helps!

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Which names are other correct names for CK Choose all answers that are correct

PilotLPTM [1.2K]

The second and third one

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3 years ago

Can someone help me asap !!

aivan3 [116]

Answer:

The answer is negative. Answer choice B

Step-by-step explanation:

Here is the question:

-3^40(5.6/2.1)

According to order of operations, we must do -3^40 first.

Its important to note that -3^40 is different than (-3)^40. In the problem you specified, the ^40 only applies to the 3, not -3. So, first we do 3^40 which is a huge number. And then, we need to add a negative symbol to that huge number. So now we have a massive negative number multiplied by 5.6/2.1, which is obviously positive. Obviously, a negative number multiplied by a positive number is still negative. Therfore, the answer is negative.

I hope my answer helped you :)

If you are happy with my answer please vote Brainliest. Thanks!

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1 year ago

A trapezoid has base lengths of 2.5 inches and 10 inches and an altitude of 8 inches. What is the area of the trapezoid?

lisov135 [29]

A= 50. 2.5+10= 12.5/2= 6.25x8= 50

5 0

3 years ago

Read 2 more answers

To prove part of the triangle midsegment theorem using the diagram, which statement must be shown

Alenkinab [10]

Answer:

The length of GH is half the length of KL.

Full question...

To prove part of the triangle midsegment theorem using the diagram, which statement must be shown?

The length of JK equals the length of JL.

The length of GH is half the length of KL.

The slope of JK equals the slope of JL.

The slope of GH is half the slope of KL.

So once again the answer is the second one: The length of GH is half the length of KL.

8 0

2 years ago

Read 2 more answers

Solve for x and y. 4x-3y=16 -2x+4y=-2 DUE 2/7/17!

Trava [24]

$\left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right.$

Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
 $\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right.$
$\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right.$
--------------------
$5y = 12$
$\boxed{\boxed{y = \frac{12}{5}}}\end{array}}\qquad\quad\checkmark$

Let's replace the value found in the first equation:
$4x-3y=16\:(I)$
$4x-3*( \frac{12}{5}) =16$
$4x - \frac{36}{5} = 16$
least common multiple (5)
$\frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5}$
$20x - 36 = 80$
$20x = 80 + 36$
$20x = 116$
$x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark$

4 0

3 years ago