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2 years ago

PLSSSSSS HELPPPPPPP FAST WILL GIVE BRAINLIEST ITS IREADY!!!!!!!!!!! PLSSSSS HELPPPPPPPPPPPPPP

Mathematics

2 answers:

Scilla [17]2 years ago

8 0

8/2 = 4 it is the ratio

soldi70 [24.7K]2 years ago

7 0

Answer:

8:1

Step-by-step explanation:

Okay so the ratio from cocoa to sugar from the chart is 8:2. To simplify this you could multiply both sides by 2 which would get you 4:1 or 4. Now the question asks what is the ratio Leah will use when this is doubled. In order to do this you can multiply 4 by 2 which gets you 8. Or 8:1. Hope you understand :)

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What is the value of X that makes equation true? -5(2x - 3) + 4x = -3x + 6

Alisiya [41]

Answer:

x=3

Step-by-step explanation:

6 0

2 years ago

What is the slope of the following equation?

777dan777 [17]

Answer: A. -3

Step-by-step explanation: The slope is the coefficent of x, this can be read as -3/1, down 3 over 1.

6 0

2 years ago

Find the Slope for <br> (3,-4) and (5,4)

REY [17]

The picture below shows the formula for calculating slope

In this example:
y2 = 4
y1 = -4
x2 = 5
x1 = 3
(These are the numbers you substitute into the formula)

Calculations:
4 - (-4) = 8
5 - 3 = 2
8/2 = 4

As you can see, after the values are substituted into the equation & simplified, the slope is equal to 4

8 0

1 year ago

Read 2 more answers

Please help :0

Zinaida [17]

Mult. f(x) = x^4 by 2 will stretch the graph vertically by a factor of 2.
Thus, eliminate (a) and (b).

Mult. x by (1/4) will stretch the graph horiz. by a factor of 4. Thus, (c) is the correct answer.

5 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago