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3 years ago

Can someone help PLEASE

Mathematics

1 answer:

Aneli [31]3 years ago

5 0

Yes . When you do the line test you go straight done and it doesn’t have 2 points on the same line .

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I need some help please

Finger [1]

Answer:

a 10000

b 1000

c 100

this is the correct answer

6 0

3 years ago

Read 2 more answers

Which value makes the equation 3x - 6 = 36 true? a) 10 b)12 c)14 d)16

ioda

Answer:

C) 14

Step-by-step explanation:

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, add 6 to both sides:

3x - 6 (+6) = 36 (+6)

3x = 36 + 6

3x = 42

Next, divide 3 from both sides:

(3x)/3 = (42)/3

x = 42/3

x = 14

c) 14 is your answer.

5 0

3 years ago

Factor this trinomial. Your answer should consist of two binomials.

Sergeu [11.5K]

Y² + 8y - 33 :

Break the expression into groups for formula ax²+ bx+c :

= (y²- 3y) + (11y - 33 )

Factor y from y² - 3y => y (y - 3)

Factor out 11 from 11 y - 33 => 11 (y - 3)

= y ( y- 3 ) + 11 ( y - 3 )

Factor out common term (y - 3 ) :

= ( y - 3 ) ( y + 11 )

hope this helps!

5 0

3 years ago

Find the # of possibilities. You are setting the combination on a four-digit lock. You want to use the numbers 1, 2, 3, and 4 bu

icang [17]

1234
1243
1432
1324
1423
1342
There's 24 different equations

4 0

3 years ago

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

6 0

3 years ago