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3 years ago

What does 3ˣ + 4·3ˣ⁺¹ = a. 13·3ˣ⁺¹ b. 5·3ˣ⁺¹ c. 5·3²ˣ⁺¹ d. 13·3ˣ

Mathematics

1 answer:

Crazy boy [7]3 years ago

3 0

Answer:

D. 13*3^x

Step-by-step explanation:

$3^x +4*3^x^+^1= \\3^x+4*3(3^x)=\\3^x(1+[4*3])=\\3^x(1+12)=\\3^x(13)=\\13*3^x$

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Lines a and b are parallel. The slope of line b is 1/3. What is the slope of line a?

Viefleur [7K]

Answer:

1/3

Step-by-step explanation:

Parallel lines have the same slope.

Therefore, if the slope of line b is 1/3, the slope of line a is also 1/3

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2 years ago

A pair of shoes cost $39 it’s on sale for 70% what is sale price? HELLPPPP

xeze [42]

Answer:

The sale price is $11.70

Step-by-step explanation:

70% of 39 is 27.3 - to find this, you just multiply 39 x .7 = 27.3

Subtract 27.3 from 39

Your get 11.7

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3 years ago

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

3 0

1 year ago

Find the exact value of each expression . Do not use a calculator .

Alex73 [517]

The attached table is one that it is convenient for you to memorize as long as you're taking courses involving angles. For the inverse functions, find the column corresponding to the function, then look for the value in that column. The angle at the beginning of that row is the answer to your question.

8. $\cos^{-1}(0)= 90^{\circ}=\dfrac{\pi}{2}$

9. $\sin^{-1}(1)= 90^{\circ}=\dfrac{\pi}{2}$

10. $\tan^{-1}\left( \dfrac{\sqrt{3}}{3} \right)=30^{\circ}=\dfrac{\pi}{6}$

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3 years ago

Triangle STU has reflected over line m to form Triangle S'T'U'. What is the value of y?

Lera25 [3.4K]

Answer:

D. 5

Step-by-step explanation:

The congruence of S'T' with ST tells you ...

9 = x +7

2 = x

The congruence of S'U' and SU tells you ...

4x = y +3

4·2 -3 = y = 5 . . . . substitute the value of x; subtract 3

3 0

3 years ago