All categories

3 years ago

Which of the following is the function for the graph? Choices/graph below.

Mathematics

1 answer:

kumpel [21]3 years ago

6 0

set x=0 to get insights.

at x=0 to the functions are

-2(0-2)²+3 = -5

-2(0+2)²-3 = -11

-2(0-2)²+3 = -5

-2(0-2)²-3 = -11

weird, for some reason option 1 and 3 seem to be the same, maybe that's an accident by the author.

maybe a + was intended inside the parentheses, dunno.

option 1 and 3 seem fine, although I think there is an error in the options. they shouldn't be exactly the same

You might be interested in

Solve the simultaneous equations by substitution<br> 4x+3y=-2<br> 3y=4-x

stiv31 [10]

Answer: x=-2 y=2

Step-by-step explanation:

I decided that isolating the y value of the second equation is the best way to start solving.

I divide 3 from both sides to get y= (4-x)/3. With this, I can substitute into the first equation. 4x+3(4-x)/3)=-2.

The multiplying the 3 by the denominator three cancels each other out to make:

4x+4-x=-2.

We subtract 4 on both sides and then subtract x from 4x.

We end up with 3x=-6. Then, we divide both sides by 3.

x=-2.

Now we substitute this into the second equation to solve.

3y=4-(-2)

This becomes 3y=4+2 which is 3y=6

We divide both sides by 3 to get y=2

We end up with x=-2 and y=2.

I hope you learned something :)

4 0

2 years ago

Read 2 more answers

First to help gets marked brainliest!!

Lubov Fominskaja [6]

Answer:

x: -8, -4, 0, 4

y: 6, 4, 2, 0

Equation: y = 2 - 0.5x

7 0

2 years ago

the domain of a function f(x) is x > 1, and the range is y < -2. What are the domain and range of its inverse function, f^

Marianna [84]

9514 1404 393

Answer:

domain: x < -2
range: y > 1

Step-by-step explanation:

Domain and range of the inverse function are swapped: the domain of the function is the range of its inverse, and vice versa.

The graph shows an example of a function and its inverse with the given domain and range. The function is red; its inverse is blue.

8 0

3 years ago

Jessie estimates the weight of her cat to be 10 pounds. The actual weight of the cat is 13.75 pounds. find the percent error.

romanna [79]

Answer:

$\large \boxed{27 \, \%} }$

Step-by-step explanation:

$\begin{array}{rcl}\text{Percent error}&= &\dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \,\%\\\\& = & \dfrac{\lvert 10 - 13.75\lvert}{13.75} \times 100 \, \% \\\\& = & \dfrac{\lvert -3.75\lvert}{13.75} \times 100 \, \%\\ \\& = & \dfrac{3.75}{13.75} \times 100 \,\%\\\\& = & 0.27 \times 100 \, \%\\& = & \mathbf{27 \, \%}\\\end{array}\\\text{The percent error is $\large \boxed{\mathbf{27 \, \%} }$}$

3 0

2 years ago

Simplify this please

Ugo [173]

Answer:

$\frac{12q^{\frac{7}{3}}}{p^{3}}$

Step-by-step explanation:

Here are some rules you need to simplify this expression:

Distribute exponents: When you raise an exponent to another exponent, you multiply the exponents together. This includes exponents that are fractions. $(a^{x})^{n} = a^{xn}$

Negative exponent rule: When an exponent is negative, you can make it positive by making the base a fraction. When the number is apart of a bigger fraction, you can move it to the other side (top/bottom). $a^{-x} = \frac{1}{a^{x}}$ , and to help with this question: $\frac{a^{-x}b}{1} = \frac{b}{a^{x}}$ .

Multiplying exponents with same base: When exponential numbers have the same base, you can combine them by adding their exponents together. $(a^{x})(a^{y}) = a^{x+y}$

Dividing exponents with same base: When exponential numbers have the same base, you can combine them by subtracting the exponents. $\frac{a^{x}}{a^{y}} = a^{x-y}$

Fractional exponents as a radical: When a number has an exponent that is a fraction, the numerator can remain the exponent, and the denominator becomes the index (example, index here ∛ is 3). $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}$

$\frac{(8p^{-6} q^{3})^{2/3}}{(27p^{3}q)^{-1/3}}$ Distribute exponent

$=\frac{8^{(2/3)}p^{(-6*2/3)}q^{(3*2/3)}}{27^{(-1/3)}p^{(3*-1/3)}q^{(-1/3)}}$ Simplify each exponent by multiplying

$=\frac{8^{(2/3)}p^{(-4)}q^{(2)}}{27^{(-1/3)}p^{(-1)}q^{(-1/3)}}$ Negative exponent rule

$=\frac{8^{(2/3)}q^{(2)}27^{(1/3)}p^{(1)}q^{(1/3)}}{p^{(4)}}$ Combine the like terms in the numerator with the base "q"

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)}q^{(1/3)}}{p^{(4)}}$ Rearranged for you to see the like terms

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)+(1/3)}}{p^{(4)}}$ Multiplying exponents with same base

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(7/3)}}{p^{(4)}}$ 2 + 1/3 = 7/3

$=\frac{\sqrt[3]{8^{2}}\sqrt[3]{27}p\sqrt[3]{q^{7}}}{p^{4}}$ Fractional exponents as radical form

$=\frac{(\sqrt[3]{64})(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Simplified cubes. Wrote brackets to lessen confusion. Notice the radical of a variable can't be simplified.