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ohaa [14]
3 years ago
14

Tina has been dieting for a total fo 13 weeks. She lost 3 pounds on the first week of her diet, but gained back a pound on the s

econd week. Ob each remaining week of here diet, sh lost 2 pounds.how many pounds has tina lost in all
Mathematics
1 answer:
PIT_PIT [208]3 years ago
6 0

Answer:

4 she lost 4 pounds, how do u not know this dood

Step-by-step explanation:


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Rochelle can run 800 meters in 5 minut
Semmy [17]
To find how much she can run in one minute we’re going to divide 800 by 5, to which we get 160.

She runs 160 meters in one minute.

We want to know how many meters she can run in 9 minutes.

160 * 9 = 1,440

She can run 1,440 meters in 9 minutes.
8 0
3 years ago
Which is the graph of y=3/4x-3<br><br> Graph A<br> Graph B<br> Graph C.
irina [24]

Answer:

Graph A

Step-by-step explanation:

I had this question in my class and got it right with answer A :')

8 0
3 years ago
1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form
lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0
3 years ago
What is the solution to the system? 1. x-y + 2 z = -7<br> 2. y + z =1<br> 3. x-2 y - 3 z = 0
kvv77 [185]

You'd find this problem easier to understand and do if you'd please list the defining equations vertically and line up variables:

1. x - 1y + 2 z = -7

2. y + 1z = 1

3. x - 2 y - 3 z = 0 Now eliminate the line numbers:

x - 1y + 2 z = -7

1y + 1 z = 1

x - 2 y - 3 z = 0

Let's use the elimination method to eliminate variable z: Seeing that z = 1 - y, we transform the first equation into 1x - 1y + 2(1-y) = -7

and the third into x - 2y - 3(1-y) = 0.

Simplifying 1x - 1y + 2(1-y) = -7

and x - 2y - 3(1-y) = 0,

we get

1x - 2y - 3 + 3y) = 0 and 1x - 1y + 2 - 2y = -7

which in turn simplify to

1x + y = 3 and 1x - 3y = -9

Having eliminated the variable z, we now focus on eliminating x. Mult. the 1st equation by -1, obtaining -1x - 1y = -3. Add this result to 1x - 3y = -9:

0 - 4y = -12, which tells us that y = 3. Subbing 3 for y in 1x + 1y = 3 tells us that x = 0.

All we have left to determine is the vaue of z.

Borrowing Equation 3, from above, we get x - 2 y - 3 z = 0, and into this equation we substitute x = 0 and y = 3: 0 -2(3) - 3z = 0.

Thus, -3z = 6, and z = -2.

The solution set is (0, 3, -2). You should check this by substitution.

3 0
4 years ago
So what is the answer 5 or 8?
Ray Of Light [21]
5 hope this help you
5 0
3 years ago
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