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Sever21 [200]
3 years ago
14

Adam purchased some items at the school store. He bought 3 pencils for $0.15 each and 5 single-subject notebook

Mathematics
2 answers:
elixir [45]3 years ago
8 0

Answer:

The answer is A

Step-by-step explanation:

Fist multiply 3 and .15 to get .45 for 3 pencils. Then subtract .45 from 6.45 and get 6. Then divided 6 by 5 to get 1.2 for each book

valkas [14]3 years ago
7 0

i think the answer is a: $1.20

my work:

3 x 0.15 = 0.45.

6.45 - 0.45 = 6.00

6.00 / 5 = 1.20

and i checked my work by doing 1.20 x 5 + 0.40 = $6.45.

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suppose that 10 is less than square rooot of n which is less than 11 what is the possible value for n
Sonja [21]

Answer:

The correct answer is that a possible value for n could be all numbers from 101 to 120.

Step-by-step explanation:

Ok, to solve this problem:

You have that: 10

Then, applying the properties of inequations, the power is raised by 2 on both sides of the inequation:

(10)^{2}

100

Then, a possible value for n could be all numbers from 101 to 120.

7 0
3 years ago
URGENT <br> WILL GIVE BRAINLIEST <br> solve using matrices<br> 4x + 5y = 40 <br> x – y = 1
dalvyx [7]

Answer:

x = 5, y = 4

Step-by-step explanation:

\left\{\begin{array}{ccc}4x+5y=40\\x-y=1\end{array}\right\\\\A=\left[\begin{array}{ccc}4&5\\1&-1\end{array}\right]\\\\detA=\left|\begin{array}{ccc}4&5\\1&-1\end{array}\right|=(4)(-1)-(1)(5)=-4-5=-9\\\\A^D=\left[\begin{array}{ccc}-1&-1\\-5&4\end{array}\right]\\\\\left(A^D\right)^T=\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]

A^{-1}=\dfrac{1}{detA}\left(A^D\right)^T\\\\A^{-1}=\dfrac{1}{-9}\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]=\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]

A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\A^{-1}A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right]

\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\left(\frac{1}{9}\right)(40)+\left(\frac{5}{9}\right)(1)\\\\\left(\frac{1}{9}\right)(40)+\left(-\frac{4}{9}\right)(1)\end{array}\right]\\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}5\\4\end{array}\right]\Rightarrow x=5,\ y=4

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Answer:

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Step-by-step explanation:

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Answer:

The area would reduce to 1/4 the original area

Step-by-step explanation:

The area of a triangle = (1/2)(base)(height) or bh/2

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Area = (1/2)(1/2 base)(1/2 height) or bh/8. Which is a reduction by 1/4

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