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3 years ago

Brainliest to the first correct answer!

Mathematics

2 answers:

Goryan [66]3 years ago

7 0

Answer:

Step-by-step explanation:

You know that 0.25 went to 2.00. That means that it's a 1.75 jump. In order to jump the other two prices up and keep the ratio the same, you must also jump them 1.75.

0.05 + 1.75 = 1.80

0.10 + 1.75 = 1.85

To double check the ration we can look back on the original. 0.05 to 0.10 is a scale of 0.05. 0.10 to 0.25 is a scale of 0.15

Then we look at our new one. 1.80 to 1.85 is a scale of 0.05. 1.85 to 2.00 is a scale of 0.15.

The ratios stayed the same, so this is the correct answer.

I hope this helped :)

Slav-nsk [51]3 years ago

5 0

D I just took a good guess and got it right

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Which technique is most appropriate to use to solve each equation drag the name of the technique into the boxing match each equa

alina1380 [7]

Answer: graphing

Step-by-step explanation:

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2 years ago

Write the equation of a line in standard form that passes through (-2, 5) and is perpendicular to

sdas [7]

Answer:

Step-by-step explanation:

x - 2y = 8

Write this in y = mx +b form

- 2y = -x + 8

Divide the equation by (-2)

$\dfrac{-2y}{-2}=\dfrac{-x}{-2}+\dfrac{8}{-2}\\\\\\y =\dfrac{1}{2}x-4$

Slope = 1/2

The slope of the perpendicular line = -1/m

$=\dfrac{-1}{\dfrac{1}{2}}=1*\dfrac{-2}{1} = -2$

m = -2 , ( -2 , 5)

Equation: y = mx +b

Plug in m = -2 ,x = -2 and y =5

5 = (-2)*(-2) + b

5 = 4 + b

5 - 4 = b

b = 1

Equation of the line:

y = mx +b

y = -2x + 1

7 0

2 years ago

What is equivalent to

eimsori [14]

Answer : The expression which is equivalent to $\sqrt[4]{9}^{\frac{1}{2}x}$ is $9^{(\frac{1}{8}x)}$ .

Step-by-step explanation :

The given expression is:

$\sqrt[4]{9}^{\frac{1}{2}x}$

As we know that:

$\sqrt[4]{x} =x^{\frac{1}{4}}$

So, by solving the above expression, we get:

$\Rightarrow 9^{(\frac{1}{4}\times \frac{1}{2}x)}$

By multiplying the power of 9, we get the following expression.

$\Rightarrow 9^{(\frac{1}{8}x)}$

Hence, the expression which is equivalent to $\sqrt[4]{9}^{\frac{1}{2}x}$ is $9^{(\frac{1}{8}x)}$ .

6 0

3 years ago

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3>Question 1</h3>

Method 1 - modelling as a quadratic with restricted domain

Assuming that the points given on the graph are points that the curve passes through, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

Equation 1 (-4, 2)

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

Equation 2 (-1, 4)

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

Equation 3 (4, 6)

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

Method 2 - modelling as a square root function

Assuming that the points given on the graph are points that the curve passes through, and the x-intercept should be included, we can model this curve as a square root function.

Given points: