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ioda
3 years ago
15

Can someone make me a

Mathematics
2 answers:
amm18123 years ago
6 0

Linear: y=3x-5

Non-Linear: y=x^2+3x+7

rjkz [21]3 years ago
6 0

linear: y=2x+1

nonlinear: y=x^2

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what is the following product of assume Y is greater than or equal to 0 square root y to the third Times Square Root why 2/3​
sattari [20]

Step-by-step explanation:

\sqrt{y3}  \times   \sqrt{y3}

\sqrt{y3 \times y3}

\sqrt{y3 + 3}

\sqrt{y6}

4 0
3 years ago
Please help me with this question!! :)
Gre4nikov [31]

Answer:

D. The last 26/8, 3-32, -Pi, -4 2/3

Step-by-step explanation:

26/8= 3.25

3-32=3.17

-Pi= -3.14

-4 2/3= -4.67

6 0
1 year ago
Find the length of a rectangular lot with a perimeter of 50 feet if the length is five more than the width.
Shalnov [3]

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syre

Step-by-step explanation:

where is the image? huh?

6 0
3 years ago
One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

Cross Multiply

9dp = c^2

or

c^2 = 9dp

Hence, the relationship between d, c and p is: c^2 = 9dp

8 0
3 years ago
1- (-6k + 6) < 3k +1 + 5k
rjkz [21]

Answer:

k>5/14

Step-by-step explanation:

4 0
3 years ago
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