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Anna35 [415]
3 years ago
15

PLEASE RESPOND IN DESCRIPTION. "Mrs. Wheeler goes to the bakery with $30. She purchases 2 loaves of bread for $3.50 each and a d

ozen donuts for
$9. She plans to spend the remainder of her money on cookies that cost $1.50 each. What is the greatest number of cookies that Mrs. Wheeler can buy?"
Mathematics
2 answers:
Advocard [28]3 years ago
6 0

Answer:

9.

Step-by-step explanation:

<em>Bread = $7</em>

<em>12 donuts = $9</em>

<em>That equals $16.</em>

<em>Cookies cost $1.50, </em>

<em>$1.50 x 9 = $13.50.</em>

AnnyKZ [126]3 years ago
5 0

Answer:

9

Step-by-step explanation:3.50

*2=7 + 9 = 16

30-16=14 1.50 * 9 =13.50$

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X-12

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Which statement is true? Millimeters are larger than centimeters. Millimeters are larger than centimeters. Meters are larger tha
jasenka [17]

Given :

Conditions :

1) Millimetres are larger than centimetres.

2) Meters are larger than kilometres.

3) Millilitres are larger than kilolitres.

4) Kilograms are larger than grams.

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Which conditions are correct.

Solution :

1) Millimetre is 0.001 m and cm is 0.01 m .

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2) Kilometer is 1000 m .

So, option 2 is also correct.

3) Millilitres is 0.001 L and kilolitres is 1000 L .

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4) Kilogram is 1000 g .

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8 0
3 years ago
Solve one, any question you want, please show work,
luda_lava [24]

Answer:

<u>Step 1:</u> Variables are q for quarters and d for dimes. So d = dimes and q = quarters

<u>Step 2:</u> The two equations that form a system of equations are:

\left \{ {{0.1d+.25q=3.70} \atop {d+q=19}} \right.

<u>Step 3: </u>Work is shown on the attachment and on the step-by-step explanation

<u>Step 4:</u> Given the context of the problem, the solution of the amount of dimes and quarters, which is 7 dimes and 12 quarters, is needed to satisfy both systems to concur that there was 7 dimes and 12 quarters in a sum of money that amounted to $3.70 with 19 coins in total.

Step-by-step explanation:

<u>Step 1:</u> To first attempt to solve this, we must have to set up two variables for the amount of value a quarter and a dime has. So since a quarter is worth 0.25¢ in and a dime is worth 0.10¢, then we can set up one of the equations to be:

0.1d + .25q = 3.70

And we set it to equal the $3.70 because that's the amount that the sum of money totals to.

<u>Step 2:</u><u> </u>Now to get the other side of the equation, we use the variables again, d and q and set that to equal 19 because that is the amount of coins that there is in total.

<u>Step 3:</u> Now that we have both equations, we can solve the problem now.

\left \{ {{0.1d+.25q=3.70} \atop {d+q=19}} \right.

Let's first solve this equation by setting one of the side's variables the same absolute value, but not the same integer value so that we can cross it off. I will choose the equation d+q=19 to be multiplied by -.1 so that we can cross off both 0.1d and -0.1d after the result of the multiplication:

\left \{ {{0.1d+.25q=3.70} \atop {-.1d+-.1q=-1.9}} \right.

Now, lets cross off everything that we can, and add up the variables together. This includes adding .25q + (-.1q) and adding 3.70 + -1.9.

Now we have:

\left \{ {{.25q=3.70} \atop {-.1q=-1.9}} \right.

Now, what we do from here is add the left side and the right side side to get one equation, which is:

.15q=1.8

Finally, we divide both sides to get q=12, which means that there are 12 quarters. Since there are 19 coins in total, we can subtract 12 from 9 to get the remaining amount of dimes, which is 7.

In conclusion, q=12 and d=7.

<u><em>My written work is also provided in the attachment!!!</em></u>

<u>Step 4:</u> What does this mean in the context of the question?

This basically means that the solution of the amount of dimes and quarters, which is 7 dimes and 12 quarters, is needed to satisfy both systems to concur that there was 7 dimes and 12 quarters in a sum of money that amounted to $3.70 with 19 coins in total.

In simpler terms, this means that we need 7 dimes and 12 quarters to satisfy that there was a sum of money that amounted to $3.70 with 19 coins in total .

3 0
3 years ago
Please help!! Im stuck on this!
Doss [256]

Since OC=8 and OC=OF, then OF=8

OEF is a right triangle, so we can find EF using

OF^2+EF^2=OE^2 \implies OF = \sqrt{OE^2-EF^2}=\sqrt{100-64}=6

Now, observe that triangles OEF and OFG are congruent: they are both right triangles, OF is common, and OE=OG because they're both radii.

So, we have

EG=EF+FG=2EF=2\cdot 6 = 12

6 0
3 years ago
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