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3 years ago

Does the number -2.675 belong to the set of integers

Mathematics

1 answer:

tankabanditka [31]3 years ago

4 0

Answer:

yes

Step-by-step explanation:

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The value -1 is a solution for x^2 - 6x + 5 = 0 True or False?

solmaris [256]

False -1 is not a solution.

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3 years ago

She mowed 10 lawns for $90. How much did she make per lawn?

Evgen [1.6K]

Answer:

9$ per lawn

Step-by-step explanation:

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3 years ago

It cost Naomi $7.70 to send 154 text messages. How many text messages did she send

s2008m [1.1K]

You have to find the unit rate

SO..

7.70 / 154 = 0.05

So it cost her 0.05 cents for each text.

So then you divide $ 5.25 by the unit rate to get the number of text

5.25 /0.05 = 105

so she sent 105 text

To check it you just multiply the unit rate by the number of text you got and you show come up with $ 5.25

0.05 x 105 = 5.25

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3 years ago

Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.

Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it $\vec{u}$ . This explanation uses both representations.

$\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right)$ .

$\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right)$ .

$\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right)$ .

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

$3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right)$ .

So is the case when the constant is negative:

$-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right)$ .

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

$3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right)$ .

Apply the two rules for the four vector operations.

$\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}$

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

$\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}$ .

$\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}$ .

7 0

3 years ago

In the figure, WY¯¯¯¯¯¯¯¯¯WY¯ and XZ¯¯¯¯¯¯¯¯XZ¯ intersect at point VV, the measure of ∠WVZ∠WVZ is (11x−19)°(11x−19)°, and the me

Zepler [3.9K]

Assuming a diagram similar to the one I've attached, ∠<em>YVZ</em> is a vertical angle to ∠<em>WVX</em>, which means they have an equal measure. Additionally, ∠<em>WVZ</em> and ∠<em>WVX</em> form a linear pair, which means they are supplementary (sum to 180°). That means we start out with the equation
$(11x-19)+(8x+28)=180$
We combine our like terms (the <em>x</em>'s get combined, then the constants get combined) and have:
$19x+9=180$
Cancel the 9 first by subtraction:
$19x+9-9=180-9 \\ 19x=171$
Cancel the 19 by division:
$\frac{19x}{19}= \frac{171}{19} \\ x=9$
Since we know that our angle we're looking for, ∠<em>YVZ</em>, is the same measure as ∠<em>WVX</em>, we substitute 9 in for <em>x</em>:
8(9)+28=72+28=100°

7 0

3 years ago