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3 years ago

What is 5/7 - 2/3 • 15/14

Mathematics

1 answer:

zimovet [89]3 years ago

7 0

5/7-2/3x 15/14
2/3x15/14=5/7
5/7-5/7=0
0

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Please help 100 points! Will mark Brainliest. Thank You!

ziro4ka [17]

HERE'S HOW TO DO IT :)

A composite figure is made up of several simple geometric figures such as triangles, rectangles, squares, circles, and semicircles.

To find the area of a composite figure, separate the figure into simpler shapes whose area can be found. Then add the areas together.

(SEE EXAMPLE ATTACHED)

HERES WHAT WE KNOW ABOUT COMPOSITE FIGURES:

Composite Figures. A figure (or shape) that can be divided into more than one of the basic figures is said to be a composite figure (or shape). For example, figure ABCD is a composite figure as it consists of two basic figures

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3 years ago

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Solve (i) 4/15+x=17/30

Anna007 [38]

Answer:

3/10

Step-by-step explanation:

4/15+x=17/30

x=17/30-4/15

x=17/30-8/30

x=9/30

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3 years ago

What is the y - intercept for y=4-2x

Alexus [3.1K]

0,4 hope that helps

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2 years ago

Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers

slamgirl [31]

Answer:

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

Step-by-step explanation:

Let be $\vec u = (2,-3)$ , its unit vector is determined by following expression:

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

Where $\|\vec u \|$ is the norm of $\vec u$ , which is found by Pythagorean Theorem:

$\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}$

$\|\vec u\| = \sqrt{13}$

Then, the unit vector is:

$\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)$

$\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$

The unit vector in component form is $\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}} \right)$ or $\hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j$ .

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3 years ago

What is 7 2/3 + 8 5/6

Archy [21]

7 2/3 + 8 5/6 = 16 1/2 ( or 32/2 )

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3 years ago

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