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3 years ago

7

¿Qué número sigue? 1; 1; 3; 15; 105;

1 answer:

Firlakuza [10]3 years ago

6 0

Answer:

307

Step-by-step explanation:

free points

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I need help don’t understand

nadya68 [22]

If f(x)=x+1 and then x became 2, you would have the function f(x)=3. So basically for that function you would be going up three over 1. That function is already g(x)=4x. If X became 2, you would have g(x)=8x. The rate of up 8 and then over one. Because of that, g(x) would be higher

8 0

3 years ago

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What is the value of y?<br><br><br><br> Enter your answer in the box.<br><br> y =

Olenka [21]

Answer:

$y=40\°$

Step-by-step explanation:

we know that

An isosceles triangle has two equal sides and two equal angles

The two equal angles are called the base angles and the third angle is called the vertex angle

In this problem the triangle ABC is an isosceles triangle

so

$AB=AC$

$5x=40\°$

$x=8\°$

The sum of the internal angles of a triangle is equal to $180\°$

so

Find the value of y

$(2y+20)\°+2(40\°)=180\°$

$2y=180\°-20\°-80\°$

$2y=80\°$

$y=40\°$

6 0

3 years ago

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15 is what percent of<br> 600?<br> Pls show work

chubhunter [2.5K]

Answer:

90

Step-by-step explanation:

600(.15) = 90

8 0

2 years ago

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What is an integer of 22,8,7,2,-11 from greatest to least

Zinaida [17]

Well, i don't know what you're asking, but -11,2,7,8,22?

3 0

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

$9dp = c^2$

or

$c^2 = 9dp$

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

8 0

3 years ago