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2 years ago

10

Please help me on this problem

2 answers:

Goshia [24]2 years ago

5 0

Answer:

Step-by-step explanation:

2•1•.25=.5

tester [92]2 years ago

3 0

Answer:

79.5 in ^3

Step-by-step explanation:

5 times 8 times 2.

40 times 2 equals 80.

2 times 1 times .25 equals .5

80-.5 equals 79.5 in^3

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What is the radius of the circle given the equation x2 - 6x + y2 + 10y + 18 = 0?

arlik [135]

$The\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-a\ coordinates\ of\ the\ center\\r-a\ radius$

$x^2-6x+y^2+10y+18=0\\\\x^2-2x\cdot3+y^2+2y\cdot5=-18\\\\x^2-2x\cdot3+3^2+y^2+2y\cdot5+5^2=-18+3^2+5^2\ \ \ |use\ (a\pm b)^2=a^2\pm2ab+b^2\\\\(x-3)^2+(y+5)^2=-18+9+25\\\\(x-3)^2+(y+5)^2=16\\\\therefore\\\\Answer:\boxed{r=\sqrt{16}=4}$

7 0

2 years ago

6. Let f and g be differentiable functions

Julli [10]

Answer:

(b) 1

Step-by-step explanation:

To differentiate $h(x)=f(x)g(x)$ we will need the product rule:

$h'(x)=f'(x)g(x)+f(x)g'(x)$ .

We have $h'(x)=f(x)g'(x)$ , so the following equation is true by the transitive property:

$f'(x)g(x)+f(x)g'(x)=f(x)g'(x)$

By subtraction property we have:

$f'(x)g(x)=0$

Since $g(x) \neq 0$ , then we can divide both sides by $g(x)$ :

$f'(x)=\frac{0}{g(x)}$

$f'(x)=0$

This implies $f(x)$ is constant.

So we have that $f(x)=c$ where $c$ is a real number.

Since $f(0)=1$ and $f(0)=c$ , then by transitive property $1=c$ .

So $f(x)=1$ .

Checking:

$h(x)=1 \cdot g(x)$

$h'(x)=g'(x)=1 \cdot g'(x)$

So the following conditions were met.

5 0

3 years ago

the local movie theatre collected a total of $8,250 last night, of which 45% came from sales at the concession stand. how much m

Wewaii [24]

45%=0.45

8,250x0.45
=3712.50

The theater collected $3712.50 at the concession stand.

3 0

2 years ago

What is X I need help i dont know how to do this

Anit [1.1K]

Answer:

x = 75

Step-by-step explanation:

Triangle BDH

The angle must add to 180 degrees

36+ 39 + DHB = 180

75 + DHB = 180

DBH = 180-75

DHB = 105

DBH + x = 180 since it is a straight line

105+ x = 180

x = 75

6 0

2 years ago

Does anyone know the the answer to the second algebra question? Will give brainiest

astraxan [27]

Answer:

$\sqrt[3]{x^{10} }[\tex]Step-by-step explanation:Exponential Rules:[tex]x^{a} + x^{b} = x^{a + b}$

$\sqrt[b]{x^{a} } =x^{\frac{a}{b} } Original Equation:[tex]\sqrt[3]{x^{10} } = x^{\frac{10}{3} } Answer:[tex]\sqrt[3]{x^{10} }[\tex]Convert the cubed root to a power. Cubed root = [tex]\frac{1}{3}$

$x^{3} x^{\frac{1}{3} }$

Convert them, so they have a common denominator - $\frac{1}{3}$

$\frac{3}{1} * \frac{1}{3}= \frac{9}{3}$

$\frac{9}{3} + \frac{1}{3} = \frac{10}{3}$

[tex]\sqrt[3]{x^{10} }[\tex] = [tex]x^{\frac{10}{3} } [\tex]

8 0

2 years ago