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2 years ago

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According to this line plot, 3 players on this team each had 15 strikeouts for a total of 45 strikeouts. How many strikeouts did

this team have in one season?

1 answer:

Masteriza [31]2 years ago

4 0

Answer: 195

Step-by-step explanation: (3x15)+(5x16)+(2x17)+(2x18)=195

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Find m/EGD. (measurement in degrees)

Wewaii [24]

Answer:

$\huge\boxed{\bf\:30^{\circ}}$

Step-by-step explanation:

We need to find m ∠EGD.

From the given figure, we can see that, ∠BGA & ∠EGD are vertically opposite angles (opposite angles that share the same vertex ⟶ G).

Also, vertically opposite angles are equal to each other.

Given, ∠BGA = 30°

So, ∠EGD = ∠BGA = 30°.

$\rule{150pt}{2pt}$

6 0

2 years ago

Read 2 more answers

Mathematics Transform Exponential Functions<br>Please answer

Vitek1552 [10]

Answer:

g(x) = 6 * 3^x

Step-by-step explanation:

This is the answer because the question states that the graph is vertically stretched by a factor of 6. Since the parent function of an exponential function is f(x) = a * 3^(k(x-d)) + c, we plug in 6 for a, where we get g(x) = 6 * 3^x.

6 0

2 years ago

Solve the problems<br><br> A) X/840 = 1/21<br><br> B) 14/X = 16/20

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

A) x/840=1/21

x=(1/21)*840=840/21=40

x=40

------------------

B) 14/x=16/20

simplify 16/20 into 4/5

14/x=4/5

x=14/(4/5)=(14/1)(5/4)=70/4

simplify

x=35/2

8 0

2 years ago

You purchase inventory from a wholesaler totaling $3,000. If the wholesaler offers you a 15/12/8 series trade discount, what is

Arada [10]

Answer: It is simple. USE YOUR BRAIN AND STOP COMING HERE TO CHEAT ON YOUR HOME WORK

Step-by-step explanation:1: GET SOME HELP

2: ASK QUESTIONS IN CLASS

3: DON'T MESS AROUND AND BE A CLOWN

4: AND THATS ON PERIOD

5 0

2 years ago

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Ad libitum [116K]

First you'll need the complementary solutions. The characteristic equation for this ODE is

$D^2+2D+1=(D+1)^2=0\implies D=-1$

with multiplicity 2. This means two linearly independent solutions will be $y_1=e^{-x}$ and $y_2=xe^{-x}$ .

Via variation of parameters, the particular solution will take the form

$y_p=y_1u_1+y_2u_2$

where

$u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}$

So, you have

$u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_1=-\displaystyle\int x\log x\,\mathrm dx$
$u_1=\dfrac14x^2-\dfrac12x^2\log x$

and

$u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_2=\displaystyle\int\log x\,\mathrm dx$
$u_2=x(\log x-1)$

So the solution to the ODE is

$y=C_1y_1+C_2y_2+y_1u_1+y_2u_2$
$y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}$
$y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)$

3 0

3 years ago