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stellarik [79]
4 years ago
12

Urgent. REALL URGENT. HHHHHHHHHHEEEEEEEEEEEEEEELLLLLLLLLLPPPPPPPPPPPPPPP MMMMMMMMMEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Mathematics
1 answer:
Effectus [21]4 years ago
8 0

Option c : 9 is the answer.

Explanation:

The expression is 1.2-(-7.8)

We need to determine the subtracted value of the expression.

From the expression we can see that the expression contains two negative signs.

To simplify the expression, we know that, When we multiply two negative numbers then the product is always positive.

This means (-) *(-) = +

Thus, the expression can be written as

1.2+7.8

Adding the two numbers, we get,

1.2+7.8=9

Thus, the value of the expression is 9.

Hence, Option c is the correct answer.

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Determine the length of the line segment. Round your answer to the nearest tenth of a unit
Paul [167]
Like of segment being r, calculated in the picture is the length

4 0
3 years ago
The two prisms (see below) are similar. The larger one has a volume of 64 cubic feet and the smaller one has a volume of 27 cubi
Brilliant_brown [7]

Answer:

7.5 feet

Step-by-step explanation:

large over small=64/27

r:4/3

4/3=10/h

4h=30

h=30/4

h=7.5 feet

7 0
4 years ago
Tomio, age 28, takes out $50,000 of straight-life insurance. His annual premium is $418.20. Using the tables found in the textbo
olya-2409 [2.1K]
<span>Tomio, age 28, takes out $50,000 of straight-life insurance. His annual premium is $418.20. Using the tables found in the textbook, determine the cash value of his policy at the end of 20 years. 

A. $26,500
B. $30,000
C. $13,250
D. $26,000

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So the cash value of this policy is $13,250
<span>
The correct answer is:
</span><span>C. $13,250</span>
6 0
4 years ago
Read 2 more answers
Hellllpppppppppppppp!!!
ANEK [815]
The answer would be 855 miles
8 0
4 years ago
Read 2 more answers
How do I find the Velocity and How long will it take for the ball to reach it's maximum height?
Colt1911 [192]
So hmm check the picture below

\bf \qquad \textit{initial velocity}\\\\&#10;h = -16t^2+v_ot+h_o \qquad \text{in feet}\\&#10;\\ &#10;\begin{cases}&#10;v_o=\textit{initial velocity of the object}\to &64\\&#10;h_o=\textit{initial height of the object}\to &12\\&#10;h=\textit{height of the object at "t" seconds} \end{cases}\\\\&#10;-----------------------------\\\\

\bf \textit{vertex of a parabola}\\ \quad \\&#10;&#10;\begin{array}{lccclll}&#10;h(t)=&-16t^2&+64t&+12\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

part 1)  

it takes  \bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds

part 2)

\bf \textit{now, doubling }v_o\\\\&#10;\begin{cases}&#10;v_o=\textit{initial velocity of the object}\to &128\\&#10;h_o=\textit{initial height of the object}\to &12\\&#10;h=\textit{height of the object at "t" seconds}\end{cases}\\\\&#10;-----------------------------\\\\&#10;\textit{vertex of a parabola}\\ \quad \\&#10;&#10;\begin{array}{lccclll}&#10;h(t)=&-16t^2&+128t&+12\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

it will reach the maximum height at   \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet


how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}} ?

subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became

4 0
3 years ago
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