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2 years ago

Is 45×18 = equal to (40×10) + (40×8) + (5 × 10 + (5×8)

Mathematics

1 answer:

astraxan [27]2 years ago

5 0

Answer:

Yes it is equal to that

Step-by-step explanation:

hope it helps

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Ms. White is going to the grocery store to buy peamut butter for $3.50 a jar. She's going to the store with

PIT_PIT [208]

Answer:

5. your question wasnt finished so there was not answer.

Step-by-step explanation:

Im guessing your question was how many jars can she by?

If so then

35 - 15 is 20

20 divided by 3.50 is 5.714

So she only has enough money left to buy 5 whole jars

4 0

3 years ago

Let L be the line with parametric equations x=5+t,y=6,z=−2−3t. Find the vector equation for a line that passes through the point

scZoUnD [109]

Answer:

The required equations are

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$ and

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$ .

Step-by-step explanation:

The given parametric equation of the line, $L$ , is $x=5+t, y=6, z=-2-3t,$ so, an arbitrary point on the line is $R(x,y,z)=R(5+t, 6, -2-3t)$

The vector equation of the line passing through the points $P(-5,7,-8)$ and $R(5+t, 6, -2-3t)$ is

$\vec P + \lambda \vec{(PR)}=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)$

The given equation can also be written as

$\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)$

The standard equation of the line passes through the point $P_0(x_0,y_0,z_0)$ and having direction $\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k$ is

$\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)$

Here, The value of the parameter, $t$ , of any point $R$ at a distance $d$ from the point, $P_0$ , can be determined by

$|t \vec v|=d\;\cdots (iv)$

Comparing equations $(ii)$ and $(iii)$

The line is passing through the point $P_0 (5,6,-2)$ having direction $\vec v=\hat i -3 \hat k$ .

Note that the point $Q(5,6,-2)$ is the same as $P_0$ obtained above.

Now, the value of the parameter, $t$ , for point $R$ at distance $d=3$ from the point $Q(5,6,-2)$ can be determined by equation $(iv)$ , we have

$|t(\hat i -3 \hat k)|=3$

$\Rightarrow t^2|(\hat i -3 \hat k)|^2=9$

$\Rightarrow 10t^2=9$

$\Rightarrow t^2=\frac {9}{10}$

$\Rightarrow t=\pm \frac {3}{\sqrt {10}}$

Put the value of $t$ in equation $(i)$ , the required equations are as follows:

For $t= \frac {3}{\sqrt {10}}$

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$

and for $t= -\frac {3}{\sqrt {10}}$ ,

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$

8 0

3 years ago

Need help with this math question pls help me will mark brainiest btw

MrRissso [65]

Answer:

Zero

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which is the hypothesis of this statement?

zubka84 [21]

I agree with you.
The hypothesis is the part where the word <em>if </em>can be found. So, the hypothesis here has to be C. a and b are negative.

3 0

3 years ago

What is one c o ck plus 2 c ock

lutik1710 [3]

Answer:

3 c ock

Step-by-step explanation:

Reason being of you add 1 c ock and 2 c ock you get 3 c ock

8 0

3 years ago

Is 45×18 = equal to (40×10) + (40×8) + (5 × 10 + (5×8)​

Is 45×18 = equal to (40×10) + (40×8) + (5 × 10 + (5×8)