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3 years ago

Jerry went to the grocery store. He bought the following items.

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

$8.22

Step-by-step explanation:

Items purchased :

Banana = 3 pounds

Price per pound of banana = $0.69

Cost of 3 pounds of banana = (3 * 0.69) = $2.07

Apple = 5 pounds

Price per pound of apple = $1.23

Cost of 5 pounds of apple = (5 * 1.23) = $6.15

Total cost :

$(2.07 + 6.15) = $8.22

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Which number line represent the solution set for the inequality 3(8-4x)<6(x-5)

quester [9]

Eliminating parentheses, you have
24 -12x < 6x -30
Adding 30 +12x, you have
54 < 18x
Dividing by 18 gives
3 < x

The appropriate number line is the one that shows the solution set as all values of x that are greater than 3.

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3 years ago

What is ⅓ of nine times a half of 10

Tems11 [23]

Answer:

Step-by-step explanation:1 third of 9 is 3. Half of 10 is 5. Multiply 5 by 3. It is now 15.

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3 years ago

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Mike drew a quadrilateral with four right angles. What shape could be drawn?

Darya [45]

There could be 2 possible shapes that could’ve been drawn.

A rectangle, or a square.

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Hope this helped!

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3 years ago

You are interested in determining which of two brands of batteries, brand A and brand B, lasts longer under differing conditions

kirza4 [7]

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3 years ago

Solve sinX+1=cos2x for interval of more or equal to 0 and less than 2pi

Igoryamba

Answer:

Question 1: $\sin(x)+1=\cos^2(x)$

Answer to Question 1: $x=0, \pi \frac{3\pi}{2}$

Question 2: $\sin(x)+1=\cos(2x)$

Answer to Question 2: $0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$

Question:

I will answer the following two questions.

Condition: $0\le x$

Question 1: $\sin(x)+1=\cos^2(x)$

Question 2: $\sin(x)+1=\cos(2x)$

Step-by-step explanation:

Question 1: $\sin(x)+1=\cos^2(x)$

Question 2: $\sin(x)+1=\cos(2x)$

Question 1:

$\sin(x)+1=\cos^2(x)$

I will use a Pythagorean Identity so that the equation is in terms of just one trig function, $\sin(x)$ .

Recall $\sin^2(x)+\cos^2(x)=1$ .

This implies that $\cos^2(x)=1-\sin^2(x)$ . To get this equation from the one above I just subtracted $\sin^2(x)$ on both sides.

So the equation we are starting with is:

$\sin(x)+1=\cos^2(x)$

I'm going to rewrite this with the Pythagorean Identity I just mentioned above:

$\sin(x)+1=1-\sin^2(x)$

This looks like a quadratic equation in terms of the variable: $\sin(x)$ .

I'm going to get everything to one side so one side is 0.

Subtracting 1 on both sides gives:

$\sin(x)+1-1=1-\sin^2(x)-1$

$\sin(x)+0=1-1-\sin^2(x)$

$\sin(x)=0-\sin^2(x)$

$\sin(x)=-\sin^2(x)$

Add $\sin^2(x)$ on both sides:

$\sin(x)+\sin^2(x)=-\sin^2(x)+\sin^2(x)$

$\sin(x)+\sin^2(x)=0$

Now the left hand side contains terms that have a common factor of $\sin(x)$ so I'm going to factor that out giving me:

$\sin(x)[1+\sin(x)]=0$

Now this equations implies the following:

$\sin(x)=0$ or $1+\sin(x)=0$

$\sin(x)=0$ when the $y$ -coordinate on the unit circle is 0. This happens at $0$ , $\pi$ , or also at $2\pi$ . We do not want to include $2\pi$ because of the given restriction $0\le x$ .