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3 years ago

Determining a Number of Solutions

Mathematics

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

Infinitely Many Solutions

Step-by-step explanation:

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Does the table represent a function?(Only 2 hours to answer!)

alisha [4.7K]

Answer: (a) Yes. The table is a function.

<u>Step-by-step explanation:</u>

In order to be a function, there cannot be any duplicate x-values.

In the given table, each x-value is different so this IS a function.

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3 years ago

The dimensions of a triangular prism are shown in the diagram.

KatRina [158]

Answer:

The answer is 192.

Step-by-step explanation:

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3 years ago

What is another way to express 42+24

Ainat [17]

Answer:

42 - (-24) = 66

Step-by-step explanation:

I'm not sure if this is the form of answer, but if you're subtracting a negative from a negative, you're adding a positive, so this is another way to write the expression.

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3 years ago

On a math test, Johnny made a raw score of 95 on a 40 question test. The correct answers were given 5 points each while incorrec

Sidana [21]

X - the number of questions he answered correctly
y - the number of questions he answered incorrectly

There were 40 questions on the test.
$x+y=40 \\ y=40-x$

The correct answers were given 5 points each, the incorrect answers were given -2 points each. He made a score of 95.
$5x-2y=95 \\
-2y=95-5x \\
y=-\frac{95}{2}+\frac{5}{2}x
$

Set 40-x and (-95/2)+(5/2)x equal to each other:
$40-x=-\frac{95}{2}+\frac{5}{2}x \\
-x-\frac{5}{2}x=-\frac{95}{2}-40 \\
-\frac{2}{2}x-\frac{5}{2}x=-\frac{95}{2}-\frac{80}{2} \\
-\frac{7}{2}x=-\frac{175}{2} \\
x=-\frac{175}{2} \times (-\frac{2}{7}) \\
x=\frac{175}{7} \\
x=25$

Johnny answered correctly 25 questions.

8 0

3 years ago

For 20 Points. 3 Qs.<br> I also gave you guys a help sheet. PLEASE HELP.

Eduardwww [97]

Wat is help sheet, didn't help,

instructions unclear, got my calculator stuck in the ceiling fan

remember some simple exponential laws
$( x^{n} )(x^{m})=x^{n+m}$ and
$\frac{x^{n}}{x^{m}} =x^{ n-m }$
$x^{-m}= \frac{1}{x^{m}}$

first one
$(2^{-6})(2^{3})=2^{-6+3}=2^{-3}= \frac{1}{2^{3}} = \frac{1}{8}$

2nd one
$\frac{7^{5}}{7^{3}} =7^{ 5-3 } = 7^{2} = 49$

third one
$(3^{-4})(3^{2})=3^{-4+2}=3^{-2}= \frac{1}{3^{2}} = \frac{1}{9}$

dunno wat to do with 'help sheet'

4 0

3 years ago