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weqwewe [10]
3 years ago
6

Help me out here guys ?

Mathematics
1 answer:
Strike441 [17]3 years ago
6 0
Hey there!

The answer to your question is D because you want to know how many tomatoes she grew
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The weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95. Round each rate to the nearest hu
stich3 [128]

Answer:

Step-by-step explanation:

Given that in the weekly ad 5 pound bag of organic apples will cost 13.95 dollars

To find unit rate

Unit rate can be found in two ways either cost of 1 apple or no of apples for 1 dollars

I part:

5 apples = 13.95 dollars

1 apple = 13.95/5 (since direct variation)

Unit rate for one apple= 2.79 dollars

Part II:

13.95 dollars = 1 apple

1 dollar = 1/13.95 = 0.071

=0.07

But normally since apples are countable as 1,2 ... we use the first type of rate only unit rate per apple

6 0
3 years ago
Can you answer these 2 questions??
Rasek [7]
Answer: no
explanation: don’t feel like it.
8 0
2 years ago
Verify the equation below with each of the values listed for z to find a solution . 3-2z=1/10
ziro4ka [17]

3-2z=\frac{1}{10} \\30-20z=1\\-20z=-29 \\z=\frac{-29}{-20} \Longrightarrow z=\boxed{1.45}

8 0
3 years ago
Read 2 more answers
What are the solutions to the quadratic equation -2x^2 + 6x + 3 = 0?
mario62 [17]

For this, we will be using the quadratic formula, which is x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}, with a=x^2 coefficient, b=x coefficient, and c = constant. Our equation will look like this: x=\frac{-6+/-\sqrt{6^2-4*(-2)*3}}{2*(-2)}


Firstly, solve the multiplications and the exponents: x=\frac{-6+/-\sqrt{36+24}}{-4}


Next, do the addition: x=\frac{-6+/-\sqrt{60}}{-4}


Next, your equation will be split into two: x=\frac{-6+\sqrt{60}}{-4},\frac{-6-\sqrt{60}}{-4} . Solve them separately, and your answer will be x=-0.436,3.436

5 0
3 years ago
Of is directly proportional toy and x = 45 when
kompoz [17]

Answer:

if x=ky, where k is constant.

value of x

45/3=k3/3

15=k, constant is 15

value of x when y =6

x=ky

x=15(6)

x=80

4 0
3 years ago
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