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3 years ago

Four consecutive odd integers sum to 304. Which integers are they?

Mathematics

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer:

i would tell you but what are the options if you tell me that ill gladly help you out

Step-by-step explanation:

would it be 25.3 or just 25

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Two times one number added to three times a second number is 21.

Daniel [21]

Answer:

Step-by-step explanation:

"Two times one number added to three times a second number is 21.

2x + 3y = 21

"The second number is ten less than five times the first number"

y = 5x-10

Substitution

2x+3(5x-10) = 21

2x + 15x - 30 = 21

17x = 51

x = 3

y = 5x-10 = 5

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3 years ago

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3 years ago

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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

$9dp = c^2$

$c^2 = 9dp$

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

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3 years ago

If f(x)=x+4 and g(x)=x^3 what is (g*f)(-3)

Molodets [167]

Step-by-step explanation:

g*f(x)=g(x+4)=(x+4)³

g*f(-3)=(-3+4)³

= 1³=1

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3 years ago

Four consecutive odd integers sum to 304. Which integers are they?​

Four consecutive odd integers sum to 304. Which integers are they?