Can someone help me please !!!!

Mathematics

1 answer:

pychu [463]3 years ago

3 0

C. I and II only
The absolute value has to be greater than negative q. And p must be greater than the absolute value of q.

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Sarah bought a $320000 house. She pays 20% down and takes out a 30-year loan for the rest. How much is the loan amount going to

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The first thing you do is mutitpy 320,000 by .20 so that you can find the amount of down payment. The down payment would be 64,000. So you subtract 64,000 from 320,000. This leaves a loan amount of 256,000 dollars. The answer is 256,000 dollars .

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What are the first 5 multiples of 150

anzhelika [568]

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Step-by-step explanation:

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Roman buys seed packets for the school garden. One packet of radish seeds costs $1.75. One packet of tomato seeds costs $2.50. L

HACTEHA [7]

Answer:

C. $t=23,r=9$ .

Step-by-step explanation:

Let r represent the cost of one packet of radish seeds and let t represent the cost of one packet of tomato seeds.

Roman spent $62.75

This implies that,

$1.75r+2.5t=62.75...eqn1$

Roman bought 32 packets of seeds and one packet of radish seeds costs
$ 1.75.
One packet of tomato seeds costs $ 2.50.

This implies that,

$r+t=32...eqn2$

Using guess and check means we substitute into the given equation.

A) We substitute $t=8,r=24$ into both equations.

$1.75(8)+2.5(24)=62.75...eqn1$

$14+60\ne 62.75...eqn1$

$8+24=32...eqn2$

Both equations are not satisfied.

B)

We substitute $t=9,r=23$ into both equations.

$1.75(9)+2.5(23)=62.75...eqn1$

$15.75+57.5\ne 62.75...eqn1$

$9+23=32...eqn2$

Both equations are not satisfied.

C)

We substitute $t=23,r=9$ into both equations.

$1.75(23)+2.5(9)=62.75...eqn1$

$40.25+22.5 =62.75...eqn1$

$23+9=32...eqn2$

Both equations are satisfied.

Hence the solution is,

$t=23,r=9$

D) We substitute $t=24,r=8$ into both equations.

$1.75(24)+2.5(8)=62.75...eqn1$

$42+20\ne 62.75...eqn1$

$24+8=32...eqn2$

Both equations are not satisfied.

4 0

3 years ago

Read 2 more answers

Find CD if C(0,3) and D(4,7)

weeeeeb [17]

$C(0,3) \\
x_1=0 \\
y_1=3 \\ \\
D(4,7) \\
x_2=4 \\
y_2=7 \\ \\
\overset{\underline{\ \ \ \ }}{CD} =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(4-0)^2+(7-3)^2}= \\
=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{16 \times 2}=4\sqrt{2} \\ \\
\boxed{\overset{\underline{\ \ \ \ }}{CD} = 4\sqrt{2}}$