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2 years ago

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Mr. Wilson drives 450 miles in seven hours. At this rate, about how many miles does he drive in six hours? A) 101 B) 254 C) 340

D) 386

1 answer:

Deffense [45]2 years ago

5 0

I think D is the answer hope this helps

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Which system of equations has infinitely many solutions?

schepotkina [342]

Answer: B.

$4x - 2y = 8\\-4x + 2y = -8$

Step-by-step explanation:

Let's solve your system by elimination.

4x−2y=8;−2x+2y=6

Add these equations to eliminate y:

4x-2y=8

-2x+2y=6

2x=14

Substitute 7 for x in 4x−2y=8:

(4)(7)−2y=8

−2y+28=8 (Simplify both sides of the equation)

−2y+28+−28=8+−28 (Add -28 to both sides)

−2y=−20

−2y/−2=−20/−2 (Divide both sides by -2)

y=10

3 0

2 years ago

Pls help pls pls pls

Romashka [77]

Answer: A don't get confused by b

Step-by-step explanation:

5 0

2 years ago

Question 9: You have 2/3 of a pumpkin pie left over from Thanksgiving. You want to give 1/2 of it to your sister. How

Sophie [7]

Answer:

1/3

Step-by-step explanation:

1/2 of 2/3 = 1/3

5 0

3 years ago

Basketball shoes are on sale for 22% off. What is the regular price if the sale price is $42?

artcher [175]

Answer:

x =53.85

Step-by-step explanation:

Let x be the original price

If the price is 22% of, you will pay 100% -22% = 78%

x * 78% = 42

Change to decimal form.

78x = 42

Divide each side by .78.

78x/.78 = 42/.78

Rounding to the nearest cent

x =53.85

4 0

2 years ago

Read 2 more answers

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago