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ella [17]
3 years ago
11

A function g(x) is defined as shown.

Mathematics
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

g(4) = 12

Step-by-step explanation:

We need to find the value the function g(x) when x = 4, that is;

g(4)

In the first definition of the function, the value of x is strictly less than 4. Consequently, we shall use the second definition of the function to evaluate g(4)

In the second definition, g(x) is given as;

Q(x) = 0.5x + 10

plug in x = 4 and simplify;

Q(4) = 0.5(4) + 10

Q(4) = 12

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is it possible to have three points that would not fit on the same plane 2 dimensional surface? Why or why not
Alisiya [41]
<span>No. Three points could be the three points of a triangle. Triangles are two dimensional surfaces. Therefore, any three points could exist on a two dimensional triangle.</span>
3 0
3 years ago
Please help! This is for Algebra 1
kow [346]
The first answer
-40y3/x5
8 0
3 years ago
Please solve this, will rate 5 stars and mark as STAR!​
Nina [5.8K]

Answer:

\boxed{5 \cdot \sqrt{2}  \cdot \sqrt[6]{5} }

Step-by-step explanation:

\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}

\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}

\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

250=2 \cdot 5^3

\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5  \sqrt[3]{2}

Once

\sqrt[6]{2}  \cdot \sqrt[6]{5} = \sqrt[6]{10}

We have

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5}

We can proceed considering the common base of exponentials

\sqrt[3]{2}  \cdot \sqrt[6]{2}  =  2^{\frac{1}{3}} \cdot  2^{\frac{1}{6} }  = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}

Therefore,

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2}  \cdot \sqrt[6]{5}

7 0
3 years ago
Which of the following are NOT congruent?
algol13

the adjacent angles in a parallelogram

5 0
3 years ago
How do you find the value of √a⁴ ??<br> (ill give brainliest)
dsp73

Answer:

a^{2}

Step-by-step explanation:

√a^4 is the same as:

√a × √a × √a × √a

group them as 2 pairs of

(√a × √a) × (√a × √a)

which makes

a × a

which is the same as a^{2}

5 0
3 years ago
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