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2 years ago

9

-10(1-9)+6(x-10) please helppppp

2 answers:

stich3 [128]2 years ago

4 0

Answer:

96x - 70

Step-by-step explanation:

Remove the brackets on both expressions.

-10 + 90x + 6x - 60

Add the xs together.

-10 + 96x - 60

combine the - 10 and - 60

96x - 70

Don't go any further. Nothing else combines.

tiny-mole [99]2 years ago

3 0

Answer:

-10 + 90x + 6x - 60

= 96x - 70

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Find the p-value in a test of the claim that the mean College Algebra final exam score of engineering majors equal to 88, give

Orlov [11]

Answer: 0.9332.

Step-by-step explanation:

Claim : College Algebra final exam score of engineering majors equal to 88.

Given that : The test statistic is z equals to 1.50.

To find the p-value (Probability value), we use standard normal distribution table, and search the p-value corresponds to the z-score.

In a Standard Normal Distribution Table below, the p-value corresponds z equals 1.5 is 0.9332.

Hence, the p-value is 0.9332.

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2 years ago

Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c).

masha68 [24]

Answer:

The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down

Step-by-step explanation:

* Lets revise the rules of the midpoint and the slope to prove the

problem

- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

* Lets solve the problem

- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)

- Lets find the mid-poits of PQ called A

∵ Point P is (x1 , y1) and point Q is (x2 , y2)

∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0

∵ A is the mid-point of PQ

∴ $A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)$

- Lets find the mid-poits of PR which called B

∵ Point P is (x1 , y1) and point R is (x2 , y2)

∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c

∵ B is the mid-point of PR

∴ $B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)$

- The parallel line have equal slopes, so lets find the slopes of AB and

QR to prove that they have same slopes then they are parallel

# Slope of AB

∵ Point A is (x1 , y1) and point B is (x2 , y2)

∵ Point A = (a , 0) and point B = (b , c)

∴ x1 = a , x2 = b and y1 = 0 and y2 = c

∴ The slope of AB is $m=\frac{c-0}{b-a}=\frac{c}{b-a}$

# Slope of QR

∵ Point Q is (x1 , y1) and point R is (x2 , y2)

∵ Point Q = (2a , 0) and point R = (2b , 2c)

∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c

∴ The slope of AB is $m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}$

∵ The slopes of AB and QR are equal

∴ AB // QR

∵ AB is the line containing the midpoints of PQ and PR of Δ PQR

∵ QR is the third side of the triangle

∴ The line containing the midpoints of two sides of a triangle is parallel

to the third side

6 0

2 years ago

A hockey player needs to shoot a puck 55 meters from his current location to his opponent's goal to score a goal. After the shot

lorasvet [3.4K]

Answer:

The puck traveled 53.8 meters.

Step-by-step explanation:

Currently, the player is 55 meters away from the opponent's goal.

After the shot, the puck is 120 centimeters = 1.2 meters from the opponents goal.

So, this means that the puck traveled 55 - 1.2 = 53.8 meters.

5 0

1 year ago

Read 2 more answers

. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of

Semenov [28]

Answer:

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$

Step-by-step explanation:

We are given that initial value problem

$t^2-20t+51)\frac{dy}{dt}=y$

$\frac{dy}{y}=\frac{dt}{t^2-20t+51}$

$\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}$

$\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}$

$\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}$

$\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}$

$\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}$

$1=A(t-17)+B(t-3)$ ...(1)

Substitute t-3=0

t=3

t-17=0

t=17

Substitute t=3 in equation (1)

$1=A(3-17)+0$

$1=-14A$

$A=-\frac{1}{14}$

Substitute t=17

$1=B(17-3)$

$1=14B$

$B=\frac{1}{14}$

Substitute the values of A and B

$\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})$

$\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}$

$ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C$

By using formula: $\frac{dx}{x}=ln x+C$

$ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C$

Using formula: $ln x-ln y=ln \frac{x}{y}$

$ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C$

$ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C$

Substitute y(10)=1

$ln 1=\frac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C$

$0=0+ln C$

Because ln 1=0

$lnC=0$

$C=e^0=1$

Because $ln x=y\implies x=e^y$

Substitute the value of C

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid$

$14ln y=ln\mid\frac{t-17}{t-3}\mid$

$lny^{14}=ln\mid\frac{t-17}{t-3}\mid$

By using identity $blog a= loga^b$

$y^{14}=\frac{t-17}{t-3}$

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$

6 0

2 years ago

Which rigid transformation would map ABC to EDC?

12345 [234]

D is the correct answer

good luck

6 0

2 years ago

Read 2 more answers