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3 years ago

Help: I am a number between 60 and 100. My ones digit is six less than my tens digit. I am a prime number.

Mathematics

2 answers:

marysya [2.9K]3 years ago

8 0

The correct answer is 97 if I’m not mistaken hope you have a wonderful day

Reptile [31]3 years ago

5 0

Answer:

Let's name the digit:

x- ones digit

y - tens digit

we know that x=y-2.

Now, y can be 6,7,8,9 (the number is between 60 and 100 (so depending on your understanding of "between", 0 is also possible. but then the number would have to have -2 as its ones digit, so in any case, it's not possible).

So the possibilities with x=y-2:

Out of those 64 and 86 are even, so they can't be prime.

75 has 5 in its ones number: it's divisible by 5.

so the correct answer is 97.

Step-by-step explanation:

Hope this helps!

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Find the perimeter of the pentagon MNPQR with vertices M(2, 4), N(5, 8), P(8, 4), Q(8, 1), and R(2, 1)

Gekata [30.6K]

Answer:

The pentagon MNPQR has a perimeter of 22 units.

Step-by-step explanation:

Geometrically speaking, the perimeter of the pentagon is the sum of the lengths of each side, that is:

$p = MN + NP + PQ + QR + RM$ (1)

$p = \sqrt{\overrightarrow{MN}\,\bullet \, \overrightarrow{MN}} + \sqrt{\overrightarrow{NP}\,\bullet \, \overrightarrow{NP}} + \sqrt{\overrightarrow{PQ}\,\bullet \, \overrightarrow{PQ}} + \sqrt{\overrightarrow{QR}\,\bullet \, \overrightarrow{QR}} + \sqrt{\overrightarrow{RM}\,\bullet \, \overrightarrow{RM}}$ (1b)

If we know that $M(x,y) = (2,4)$ , $N(x,y) = (5,8)$ , $P(x,y) = (8,4)$ , $Q(x,y) = (8,1)$ and $R(x,y) = (2,1)$ , then the perimeter of the pentagon MNPQR is:

$p =\sqrt{(5-2)^{2}+(8-4)^{2}} + \sqrt{(8-5)^{2}+(4-8)^{2}}+\sqrt{(8-8)^{2}+(1-4)^{2}}+\sqrt{(2-8)^{2}+(1-1)^{2}}+\sqrt{(2-2)^{2}+(4-1)^{2}}$ $p = \sqrt{3^{2}+4^{2}} + \sqrt{3^{2}+(-4)^{2}}+\sqrt{0^{2}+(-3)^{2}}+\sqrt{(-6)^{2}+0^{2}}+\sqrt{0^{2}+3^{2}}$

$p = 22$

The pentagon MNPQR has a perimeter of 22 units.

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3 years ago