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2 years ago

Need help on this math problem

Mathematics

1 answer:

erik [133]2 years ago

7 0

Answer:

$-5, 18, \sqrt{13}$

Step-by-step explanation:

We can solve the first equation, f of -3. The value of the function f is $\frac{1+x^2}{x+1}$ , and plugging in -3 gets us $\frac{1+9}{1-3}$ , this results in 10 divided by negative 2, which is negative 5.

Now, we must solve g of negative one third. The function g is defined as $|9x-15|$ . Plugging in negative one third into the question gets us $|9(-\frac{1}{3})-15|$

9 times negative one third is -3, and -3 minus 15 is -18. The absolute value of -18 is 18.

Now, we must solve h of negative 2, and h is defined as $\sqrt{-3-8x}$ . Plugging in negative 2, we have $\sqrt{-3-8(-2)}$ . Negative 8 times negative 2 is positive 16, and 16 minus 3 is 13. The answer is the square root of 13

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Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

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3 years ago

Find the limit of the function algebraically. limit as x approaches negative nine of quantity x squared minus eighty one divided

scoray [572]

Let me express the equation clearly:

lim x→-9 (x²-81)/(x+9)

Initially, we solve this by substituting x=-9 to the equation.

((-9)²-81)/(-9+9) = 0/0

The term 0/0 is undefined. This means that the solution is not see on the number line because it is imaginary. Other undefined terms are N/0 (where N is any number), 0⁰, 0×∞, ∞-∞, 1^∞ and ∞/∞. One way to solve this is by applying L'Hopitals Rule. This can be done by differentiating the numerator and denominator of the fraction independently. Then, you can already substitute the x=-9.

(2x-0)/(1+0) = 2x = 2(-9) = -18

The other easy way is to substitute x=-8.999 to the original equation. Note that the term x→-9 means that x only approaches to -9. Thus, you substitute a number that is very close to -9. Substituting x=-8.999

((-8.999)²-81)/(-8.999+9) = -18

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3 years ago

Joanne has a cylindrical, above ground pool. the depth (height) of the pool is 1/2 of its radius, and the volume is 1570 cubic f

diamong [38]

Answer:

314 ft²

Step-by-step explanation:

Data provided in the question:

Depth (height) of the pool = $\frac{1}{2}$ of its radius

let the radius be 'r' ft

thus,

Depth (height) of the pool, h = $\frac{r}{2}$ ft

Volume of the pool = 1570 ft³

also,

Volume of cylinder = πr²h

thus,

1570 ft³ = πr²h

1570 ft³ = $\pi r^2(\frac{r}{2})$

3140 = 3.14 × r³ [ π = 3.14 ]

r³ = 1000 ft³

r = 10 ft

Thus,

Area of the bottom surface = πr²

= 3.14 × 10²

= 314 ft²

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3 years ago

. Find the simple interest on a $2,219.00 principal, deposited for 6 years at a

koban [17]

Answer:

SI=PTR/100

SI=2129*6*1.91/100

SI=24398.34/100

SI=243.9834$

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3 years ago

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liubo4ka [24]

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3 years ago

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Need help on this math problem​

Need help on this math problem