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2 years ago

Abigail has a bag that contains pineapple chews, lemon chews, and watermelon

Mathematics

1 answer:

musickatia [10]2 years ago

6 0

Answer:

w=8

w+p+l = 35

=8\35 =0.228

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If a ÷ b = 2 and b ÷ c =3/4, what is the value of c ÷ a? Express your answer as a common fraction.

lora16 [44]

Answer:

in the picture above.

Step-by-step explanation:

I hope that it's a correct answer.

7 0

2 years ago

Jamarius buys T-shirts for $6 each and marks up the price by 45%. How much profit does Jamarius make from each T-shirt sold?

3241004551 [841]

Answer:

$2.70

Step-by-step explanation:

To add 45 percent to the original price you do 100%+45%=145%=1.45

times 1.45 by the original price.

Then take away the original price from the answer you just got.

$1.45×$6=$8.70

$8.70-$6=$2.70

6 0

2 years ago

What is the following product?<br> 3 Square root of four times square root of three

dolphi86 [110]

Answer:

the answer is option B.

$\sqrt[3]{12}$

8 0

3 years ago

HELP PLEASEEEEEEEEEEEEEE

vazorg [7]

Answer:

-2/4, 8/40, 21/30, 9, 9 20/25

Step-by-step explanation:

9/20/25 is a bit greater than 9.

21/30 is 9 away from a whole 8/40 is 32 away from 40 -2/4 is way less because its on the negative side which is like so far away from the other numbers.

7 0

2 years ago

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

2 years ago