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3 years ago

Point Q is on line segment PR. Given QR=11 and PQ=3, determine the length PR.

Mathematics

2 answers:

klio [65]3 years ago

3 0

Answer:

$\huge \boxed{14}$

Step-by-step explanation:

Q is a point on the line segment PR.

PQ = 3

QR = 11

PR = PQ + QR

PR = 3 + 11 = 14

Anna11 [10]3 years ago

3 0

Answer:

$\huge\boxed{PR = 14}$

Step-by-step explanation:

QR = 11

PQ = 3

Given that Q is on line segmant PR

So,

PR = PQ + QR

PR = 3 + 11

PR = 14

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AVprozaik [17]

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A.i. the circumference of a circle is nothing else but its perimeter which formula is π × diameter = π × 2 × radius

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3 years ago

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F(x)=1/2x+3; Find f(8)

Shtirlitz [24]

Answer:

Step-by-step explanation:

1/2(8)+3

4+3

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4 years ago

Divide. 413÷129 Enter your answer, as a mixed number in simplest form, in the box.

kompoz [17]

For this case we must divide the following expression:

$\frac {413} {129}$

The expression can not be simplified. When dividing we have that its decimal form is given by:

3,201550

If we convert to a mixed number we have:

$3 \frac {26} {129}$

Verifying:

$3 \frac {26} {129} = \frac {129 * 3 + 26} {129} = \frac {413} {129}$

Answer:

$3 \frac {26} {129}$

5 0

3 years ago

What is the greatest common factor of the terms of the polynomial 12x^4 - 6x^3 + 9x^2

choli [55]

Answer:

$3x^{2}$ is the greatest common factor of the polynomial.

Step-by-step explanation:

Given:

The polynomial is $12x^{4} - 6x^{3} + 9x^{2}$

In order to determine the greatest common factor of all the terms, we find the greatest common factor of the coefficients separately and greatest common factor of the variables separately.

So, factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 6 = 1, 2, 3, 6

Factors of 9 = 1, 3, 9

Therefore, the greatest common factor among 12, 6, and 9 is 3.

Now, factors of $x^{4}$ = $1, x, x^{2}, x^{3}, x^{4}$

Factors of $x^{3}$ = $1, x, x^{2}, x^{3}$

Factors of $x^{2}$ = $1, x, x^{2}$

Therefore, the greatest common factor among $x^{4}, x^{3},x^{2}$ is $x^{2}$ .

Hence, the greatest common factor of all the terms of the polynomial is $3x^{2}$

3 0

4 years ago

The slope at any point on the graph of the function h(x) where h(1)=9 is given by the expression dy/dx=<img src="https://tex.z-d

Gekata [30.6K]

The given differential equation is separable:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\sqrt y\implies\dfrac{\mathrm dy}{\sqrt y}=2x\,\mathrm dx$

Integrate both sides:

$2\sqrt y=x^2+C$

In this case, $y=h(x)$ , so that given $h(1)=9$ , we get

$2\sqrt9=1^2+C\implies C=5$

so that

$2\sqrt{h(x)}=x^2+5\implies h(x)=\left(\dfrac{x^2+5}2\right)^2=\dfrac{x^4+10x^2+25}4$

7 0

3 years ago