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kirill115 [55]
3 years ago
8

List the integers that satisfy both these inequalities. Put the answers on one line.

Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0

Answer:

-7, -6, -5, -4, -3

Step-by-step explanation:

Given:

2x + 5 < 0

x > -8

Thus, we can combine both statements to find out integers that satisfy both inequalities.

2x + 5 < 0

Let's find x

2x < 0 - 5 (substraction property)

2x < -5

Divide both sides by 2

x < -5/2

This implies that -5/2 is greater than the set of values of x.

The second inequality, x > -8 implies that -8 is less than the value of x.

Les combine both:

-8 < x < -5/2

Therefore, the possible set of integers are whole numbers between the range of -8 and -5/2 which excludes -8 and -5/2.

Thus, they are:

-7, -6, -5, -4, -3

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nikitadnepr [17]
4.5 • 10-5 = 40

2.4 • 10-2 = 22
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I need help with this
Ostrovityanka [42]

Answer:

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Step-by-step explanation:

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4 years ago
If y-3x=9, 7x+y=25, what is the value of x and y?
lianna [129]

\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

4 0
2 years ago
Whoever has the best answer gets the brainliest answer!:)
Korvikt [17]
Since the length is 10 in. You must multiply it by \frac{1}{5} and then multiply by \frac{1}{2}. The reason for this is because if you divide ten by 5 it will be 2, then if you divide 2 by 2 it will be 1.
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4 years ago
Polygon q is a scaled copy of polygon p using a scale factor of 1/2 polygon q’s area is what fraction of polygon p’s area?
san4es73 [151]

Answer:

Polygon q’s area is one fourth of polygon p’s area

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z-----> the scale factor

x-----> polygon q’s area

y-----> polygon p’s area

so

z^{2} =\frac{x}{y}

In this problem we have

z=\frac{1}{2}

substitute

(\frac{1}{2})^{2} =\frac{x}{y}

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x=\frac{1}{4}y

therefore

Polygon q’s area is one fourth of polygon p’s area

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3 years ago
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