All categories

2 years ago

I need help with this one can anyone help ?

Mathematics

1 answer:

Novosadov [1.4K]2 years ago

6 0

(F) 0.2

10/50 = 1/5 = 0.2

(G) 1.6667

5/3 = 1 + (2/3) = 1 + 0.666....

(H) 1.11394

Use a calculator

(I) DNE

Dividing by 0 is not allowed.

You might be interested in

What is the range of the function

fiasKO [112]

Answer:

Step-by-step explanation:

6 0

3 years ago

Circle the expression(s) that give the same product as 6 x 3/8.

34kurt

Answer:

(6 × 3) divided by 8 and 3/8 × 6

Step-by-step explanation:

if you solve 6 × 3/8 you should get 18/8 which then simplifies to 9/4. if you go though each answer choice solving each problem you should get 9/4 and if you dont then that is not the answer. the first 2 answer choices equal 16 which is not 9/4 and the 4th answer choice will give you 4 and 6 × 8/3 gives you 48/3 which gives you 16 so only the 3rd and last answer choices are correct

7 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!! 20 points

irina [24]

Answer:

if they are same exact thing but mirrored, then just mirror it the way it needs to.

Example

-1 in the second by -3.4, 3.4 by 1

5 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

4 years ago

Use substitution X+y=11 -x=-y-9

Alborosie

Answer:

-(_y)-11=-y-9

Y-11=-y-9

2y=2

Y=1

4 0

3 years ago