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3 years ago

I need help, can someone help

Mathematics

1 answer:

saw5 [17]3 years ago

7 0

The answers on the right are in reverse order to the questions on the left.

I find it easier to see if I have a picture.

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A hiker walks at a speed of 3 2/5 kilometers per hour. How many kilometers will the hiker walk-in 3/4 hours?

kramer

Answer: 2.55 (or 51/20) kilometers

Step-by-step explanation: 17/5 km per one hour is 51/20 (2.55) km per 3/4 hours

6 0

2 years ago

The first car of the D-line train carried 1.5 times the number of passengers that the second car carried. Three passengers get o

kondor19780726 [428]

Answer:

the first car has 15 passengers and the second has 10

Step-by-step explanation:

7 0

3 years ago

Find the radius of a circle whose circumference is 31.4 centimeters. Approximate Π as 3.14.

AnnyKZ [126]

Circumference is Diameter * 3.14

so D * 3.14 = 31.4

31.4 / 3.14 = 10 so the diameter is 10.

the radius is half of the diameter so radius = 5 centimeters.

4 0

2 years ago

Read 2 more answers

PLEASE HELP> I NEED ANSWER ASAP. GEOMETRY

OlgaM077 [116]

Answer:

Angle 4=82 degrees, then i think angle 1 and 3= 98 degrees

Step-by-step explanation:

Angle 2=Angle 4 and 180-82=98

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

6 0

3 years ago