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3 years ago

Can someone help me with this one, please

Mathematics

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

<h2>Sorry I couldn't find it.</h2>

Step-by-step explanation:

You might be interested in

Convert 2000 ft./min. to miles per hour round to the nearest 10th

Serggg [28]

Answer:

The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour .

Step-by-step explanation:

Given as :

The speed = s = 2000 ft per min

Now, ∵ 5280 feet = 1 miles

so, 1 feet = $\dfrac{1}{5280}$ miles

∴ 2000 feet = $\dfrac{1}{5280}$ × 2000 miles

i.e 2000 feet = 0.378 miles

<u>Again</u>

∵ 1 minute = $\dfrac{1}{60}$ hour

So, The speed in miles per hour

s = $\dfrac{distance}{time}$

∴ s = $\frac{0.378}{\frac{1}{60}}$

i.e s = 2.268 miles per hour

Hence, The conversion of 2000 ft per min into miles per hour is 2.268 miles per hour . answer

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

1. If h(x) = 4x - 2, then which of the following is the solution to h(x) = 3 ?

luda_lava [24]

Answer:

Step-by-step explanation:

h(x)=4x-2 h(x)=3

h(3)=4(3)-2 substitute

h(3)=12-2 multiply

=10 subtract

8 0

3 years ago

Can someone help me with 1-5 please because i’m really struggling

viva [34]

9514 1404 393

Answer:

angle SVR = 30°
angle PVR = 90°
angle SVQ = 60°
angle SVP = 120°
angle RVQ = 30°

Step-by-step explanation:

As with finding distances on a number line, the easiest way is to subtract the smaller coordinate from the larger. Here, the "number line" is the arc at the edge of the protractor. The corresponding coordinates are ...

S: 40°

R: 70°

Q: 100°

P: 160°

The measure of an angle with its vertex at V will be the difference between the "coordinates" of the endpoints of the arc it intercepts. That is ...

m∠SVR = "R" - "S" = 70° -40° = 30°

The other angles are found the same way:

m∠PVR = 160° -70° = 90°

m∠SVQ = 100° -40° = 60°

m∠SVP = 160° -40° = 120°

m∠RVQ = 100° -70° = 30°

3 0

3 years ago

Need help with mymathlab problem

Goshia [24]

I can’t see your math lab problem, it’s not possible unless you could have shown me the footage based on the problem

7 0

3 years ago

Read 2 more answers