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mash [69]
3 years ago
5

The value of 8 is between which two integers? 2 and 3 3 and 4 07 and 9 49 and 81

Mathematics
1 answer:
Delvig [45]3 years ago
5 0

Answer:

2 and 3

Step-by-step explanation:

The square root of 8 is 2.82842712475

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Use the distributive property to simplify the following expression: -3(-x - 3y)
alexandr402 [8]

Step-by-step explanation:

-3(-x -3y) = 3( x +3y) = 3x+9y

7 0
3 years ago
Equation editor does not include the grouping symbols “<“ and “>” that are necessary for writing a vector in component for
Scilla [17]

Answer:

{12,2}

Step-by-step explanation:

From the given graph it is clear that the initial point of the vector is (-5,0) and the terminal point (7,2).

If initial point of a vector is (x_1,y_1) and terminal point is (x_2,y_2), then

\overrightarrow{v}=(x_2-x_1)i+(y_2-y_1)j

Using this formula, we get

\overrightarrow{v}=(7-(-5))i+(2-0)j

\overrightarrow{v}=12i+2j

\overrightarrow{v}=

Using braces, we get

\overrightarrow{v}=\{12,2\}

Therefore, the required vector is {12,2}.

4 0
3 years ago
Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a
sdas [7]

Answer:

-15

Step-by-step explanation:

We proceed as follows;

In this question, we want to fill in the blank so that we can have the resulting expression expressed as the product of two different linear expressions.

Now, what to do here is that, when we factor the first two expressions, we need the same kind of expression to be present in the second bracket.

Thus, we have;

2a(b-3) + 5b + _

Now, putting -15 will give us the same expression in the first bracket and this gives us the following;

2a(b-3) + 5b-15

2a(b-3) + 5(b-3)

So we can have ; (2a+5)(b-3)

Hence the constant used is -15

8 0
3 years ago
A segment in the complex plane has a midpoint at –1 + i. If the segment has an endpoint at –5 – 7i, what is the other endpoint?
morpeh [17]

Answer:

The other end point is: s+ti = 3+9i

Step-by-step explanation:

Mid-Point(M) in the complex plane states that the midpoint of the line segment joining two complex numbers a+bi and s+ti is the  average of the numbers at the endpoints.

It is given by:    M = \frac{a+s}{2} +(\frac{b+t}{2})i

Given: The midpoint = -1 + i and the segment has an endpoint at -5 - 7i

Find the other endpoints.

Let a + bi = -5 -7i  and let other endpoint s + ti (i represents imaginary )

Here, a = -5 and b = -7 to find s and t.

then;

-1+i = \frac{-5+s}{2} + ( \frac{-7+t}{2})i     [Apply Mid-point formula]

On comparing both sides

we get;

-1 = \frac{-5+s}{2}  and  1 = \frac{-7+t}{2}

To solve for s:

-1 = \frac{-5+s}{2}

or

-2 = -5+s

Add 5 to both side we have;

-2+5 = -5+s+5

Simplify:

3 = s or

s =3

Now, to solve for t;

1 = \frac{-7+t}{2}

2 =-7+t

Add 7 to both sides we get;

2+7 = -7+t+7

Simplify:

9 = t

or

t =9

Therefore, the other end point (s+ti) is, 3+9i




4 0
4 years ago
Read 2 more answers
Pleasee i need help very bad
Anon25 [30]

Angles UST and TSU are equivalent.

The answer is TSU

3 0
3 years ago
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