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REY [17]
2 years ago
10

Find g(-3) g(2) g(4)

Mathematics
1 answer:
Stells [14]2 years ago
4 0

Answer:

\frac{1}{2} , 2 , 4

Step-by-step explanation:

x = - 3 ≠ 2 , then

g(- 3) = \frac{1}{2} × (- 3)² - 4 = \frac{1}{2} × 9 - 4 = \frac{9}{2} - 4 = \frac{1}{2}

x = 2 then g(2) = 2

x = 4 ≠ 2 , then

g(4) = \frac{1}{2} × 4² - 4 = \frac{1}{2} × 16 - 4 = 8 - 4 = 4

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the difference of two numbers is eight the first number x is one less than twice the second number y write and solve a system of
jeka94
<h3>Answer:</h3>

The two numbers are 17 and 9.

<h3>Explanation:</h3>

<em>System of Equations</em>

Let x and y represent the first and second numbers, respectively.

... x - y = 8 . . . . . . . . . the difference of the two numbers is 8

... x = 2y - 1 . . . . . . . . .the first number is one less than twice the second

<em>Solution</em>

The second equation gives an expression for x that can be substituted into the first equation.

... (2y -1) -y = 8

... y -1 = 8 . . . . . . . collect terms

... y = 9 . . . . . . . . . add 1

... x = 2·9 -1 = 18 -1

... x = 17

The first number is 17; the second number is 9.

4 0
4 years ago
Solve for x.<br> 6(х – 1) = 9(х + 2)<br> Ox=-8<br> 0 x = -3<br> Ox= 3<br> Ox= 8
Maru [420]
The answer to you’re question would be A. x = -8
3 0
3 years ago
Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)
Drupady [299]

Answer:

The equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

Let us solve the expression step by step to get the equivalent

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

as

\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}      ∵ \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=x^{2\cdot \frac{1}{7}}

=x^{\frac{2}{7}}

also

\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}         ∵  \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=y^{3\cdot \frac{1}{5}}

=y^{\frac{3}{5}}

so the expression becomes

\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)            ∵ \:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}

Thus, the equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

5 0
3 years ago
Find the remainder , when p(x)=x^2-3x+6 is divided by (x-1)
OlgaM077 [116]
Pretty sure it’s 4/x-1 if u use synthetic division.
6 0
3 years ago
29<br> turn a fraction to a decimal
densk [106]
Divide the numerator by the denominator
3 0
4 years ago
Read 2 more answers
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