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2 years ago

18.2 + 45.4 + x < 100

2 answers:

8 0

Simplifying 18.2 + 45.4 + x = 100
Combine like terms: 18.2 + 45.4 = 63.6 63.6 + x = 100
Solving 63.6 + x = 100
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-63.6' to each side of the equation. 63.6 + -63.6 + x = 100 + -63.6 Combine like terms: 63.6 + -63.6 = 0.0 0.0 + x = 100 + -63.6 x = 100 + -63.6 Combine like terms: 100 + -63.6 = 36.4 x = 36.4
Simplifying x = 36.4

NemiM [27]2 years ago

4 0

18.2+45.4=63.6
100-63.6=36.4

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2 years ago

NEED HELP ASAP

mario62 [17]

1) (-7)3

2) PEMDAS; simplify (5-3)

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3 years ago

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Solve for z: 25z+85=90 z= this confuses me and I need help ;-;

seraphim [82]

Z=1/5

Subtract 85 from both sides which gets you 25z=5.

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2 years ago

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What is the answr to negative 5 to the third power times negative one

tia_tia [17]

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6 0

2 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 $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

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3 years ago