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3 years ago

Easy math question Help this is due today

Mathematics

1 answer:

Natasha2012 [34]3 years ago

7 0

Answer:

Use the pythagorean theorem to solve.

Step-by-step explanation:

$2^{2}+3^{2}=c^2\\4+9=c^2\\13=c^2\\\\\\sqrt{13} =\sqrt{c^2}$

The square root and the square cancel out. Then your answer will be the square root of 13 or 3.60555127546. Since the instructions say to round to 2 places we will do so. 3.61 meters is the length of the slide

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Answer the right question

SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

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3 years ago

Jon has $45.00. He plans to spend 4/5 of his money on sports equipment. How much will <br> h spend?

Vika [28.1K]

For this, you simply divide 45/5 which gets you 9 and you multiply that by 4 which gets you 36$

6 0

3 years ago

Read 2 more answers

What is the radius and diameter of the following circle?<br> 11 cm

yanalaym [24]

Answer:

Diameter is 11cm, and the radius is 5.5

Step-by-step explanation:

To get the radius, just divide the size of the circle by 2

Diameter is the same size as the circle

3 0

3 years ago

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

nika2105 [10]

Answer:

The components of $\vec{b}$ parallel and perpendicular to $\vec {a}$ are $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ and $\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$ , respectively.

Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$ , the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$ , dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$ , whose value is determined by Pythagorean Theorem.

The component of $\vec{b}$ parallel to $\vec {a}$ is:

$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

$\|\vec {a}\| = \sqrt{10}$

$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

$\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j$

$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

If $\vec b = 2\,i+j-3\,k$ and $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ , then:

$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$

4 0

3 years ago

State all integer values of x in the interval (-3, 3] that satisfy the following

Galina-37 [17]

(-3,3] is an interval which means that -3 and 3 are not inclusive of the values of x -5/3.

We have given

-3x+4 > 9

Required values of x.

State the values of x within -3 and 3.

subtract 4 from both sides,

-3x+4-4>9-4

<h3>What is the meaning of like terms?</h3>

like terms are terms that have the same variables and powers. The coefficients do not need to match.

Collect Like Terms

-3x > 5

(-3x)(-1) < 5(-1)

3x < -5

Divide both sides by 3

3x/3 < -5/3

x < -5/3

(-3,3] is an interval which means that -3 and 3 are not inclusive of the values of x is -5/3.

To learn more about the interval visit:

brainly.com/question/12221823

#SPJ1

3 0

3 years ago