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3 years ago

Can someone help me pls

Mathematics

1 answer:

Ivenika [448]3 years ago

7 0

Answer:

(-0.5, 0)

Step-by-step explanation:

The easiest way to do this question is by going to the left 0.5 and seeing how far you need to go up or down.

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A typical tip in a restaurant is 15% of the total bill. If the bill is $160, what would the typical tip be?

Westkost [7]

Answer:

24$

Step-by-step explanation:

15%=0.15

160*0.15=24

4 0

2 years ago

Variable y varies directly with x2, and y = 96 when x = 4.

lakkis [162]

Direct variation is y = kx, where k is the constant of variation.
But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.

The equation is y = k(2x)
Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12

The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made. y = 12(2*2) y = 48
_________________
Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2

But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.

Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.

4 0

3 years ago

Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

$k=0$

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

$\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}$

Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

3 0

2 years ago

Please Help! I need help to my math question

dem82 [27]

First you need to get rid of the parenthesis by distributing the 0.6 to each term inside.
(0.6)(10n) + (0.6)(25) =10+5n
6n +15 = 10+5n subtract the 5n on both sides and subtract 15 from both sides
6n-5n = 10-15
n=-5

5 0

3 years ago

Read 2 more answers

An airplane 40,000 feet above the ground begins descending at a rate of 1,500 feet per minute. Write an equation to model the si

user100 [1]

Answer:

h(t) = 40,000 - 1,500t

Step-by-step explanation:

Let h(t) be the height of the airplane after t minutes.

h(t) = 40,000 - 1,500t

4 0

3 years ago