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Evgen [1.6K]
3 years ago
9

Which inequalities would have a closed circle when graphed? Check all that apply.

Mathematics
2 answers:
Vesna [10]3 years ago
8 0

Answer:

Step-by-step explanation: Which inequalities would have a closed circle when graphed? Check all that apply.

x > 2.3

5.7 Less-than-or-equal-to p

One-half greater-than y

m Greater-than-or-equal-to 10

s < –7.6

The answer is B and D

Fudgin [204]3 years ago
3 0

Answer:

5.7 < p and m > 10

Step-by-step explanation:

It is what it is.

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The length of a rectangular is 7/10 ft .and the width is 1/5ft what is the perimeter
Blababa [14]
<h3>Answer:</h3>

\huge\boxed{1 \frac{4}{5}\ \text{ft}}

<h3>Explanation:</h3>

Change \frac{1}{5} so that is has a common denominator.  \frac{1*2}{5*2}=\frac{2}{10}

Make an equation.  \frac{7+7+2+2}{10}

Add.  \frac{14+4}{10}

Add.  \frac{18}{10}

Divide both sides by 2.  \frac{9}{5}

Convert to a mixed number.  1 \frac{4}{5}

4 0
3 years ago
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The width of the rectangle is 75% of its length, the length is 24, what's the area?
lapo4ka [179]
S = a * b  where a - <span>length and b - width
a = 24
b = 0.75 * a
S = 24 * 24 * 0.75 = 432</span>
7 0
3 years ago
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Help please . The answer choices are <br> A. 0 <br> B. 1<br> C. 7<br> D. Undefined
frosja888 [35]

The answer would be C

5 0
3 years ago
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Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
HELPPP I NEED TO DO THIS BEFORE 7:00​
Oksana_A [137]

Answer:

A, (13/3 x 5/13)

Step-by-step explanation:

So first, you should change the mixed numbers into improper fractions.  To convert, just multiply the whole number by the denominator and add the numerator.  Do that to both fractions and you should get 13/3 ÷ 13/5.  Since you need to switch to multiplication, switch the sign and flip that fraction/  That asnwer is 13/3 x 5/13.  Hope that helps!  :)

3 0
3 years ago
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