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timama [110]
3 years ago
5

A wildlife had 8 elephant calves born during the summer and now has 31 total elephants. How many elephants were in the reserves

before the summer?
Mathematics
2 answers:
seraphim [82]3 years ago
6 0
The answer should be 23 elephant calves
shutvik [7]3 years ago
6 0
Before the summer there was 23 elephants. This is why.
31 total elephants minus the newest 8 calves equals 23. This makes 23 the amount of elephants before summer. Hope I assisted!
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Please help and thank you
tatiyna

Answer:

b_{1}=\frac{2A}{h}-b_{2}

Step-by-step explanation:

Given: A=\frac{1}{2}(b_{1} +b_{2})h

We need to completely isolate b_{1} to solve.

A=\frac{1}{2}(b_{1} +b_{2})h

A=(\frac{1}{2}b_{1} +\frac{1}{2} b_{2})h

A=\frac{1}{2}b_{1} h+\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}h+A=\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}h=-A+\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}=\frac{-A}{h}+\frac{1}{2}b_{2}

Finally, multiply both sides by -2 to completely isolate b_{1}.

b_{1}=\frac{2A}{h}-b_{2}

5 0
3 years ago
Read 2 more answers
ifying an Error Examine the work shown. Explain the error and find the correct result. 2(4 – 16) – (–30) 2(–12) – (–30) 24 – (–3
Dvinal [7]

Answer:

The error is in the middle: 2(-12) = -24 not 24.

Step-by-step explanation:

7 0
3 years ago
Read 3 more answers
How do i resolve this?
EleoNora [17]
Solve for y for both equations then equate and solve:

5x + 9y = 160
9y = 160 - 5x
y = 160/9 - 5/9x

9x + 8y = 202
8y = 202 - 9x
y = 202/8 - 9/8x

160/9 - 5/9x = 202/8 - 9/8x (common denominator is 72)

1280/72 - 40/72x = 1818/72 - 81/72x
1818/72 - 1280/72 = 81/72x - 40/72x
538/72 = 41/72x
538 = 41x
538/41 = x
x = 13 5/41
3 0
3 years ago
Given $m\geq 2$, denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod
goblinko [34]

(2+3)^{-1}\equiv5^{-1}\pmod7 is the number <em>L</em> such that

5L\equiv1\pmod7

Consider the first 7 multiples of 5:

5, 10, 15, 20, 25, 30, 35

Taken mod 7, these are equivalent to

5, 3, 1, 6, 4, 2, 0

This tells us that 3 is the inverse of 5 mod 7, so <em>L</em> = 3.

Similarly, compute the inverses modulo 7 of 2 and 3:

2a\equiv1\pmod7\implies a\equiv4\pmod7

since 2*4 = 8, whose residue is 1 mod 7;

3b\equiv1\pmod7\implies b\equiv5\pmod7

which we got for free by finding the inverse of 5 earlier. So

2^{-1}+3^{-1}\equiv4+5\equiv9\equiv2\pmod7

and so <em>R</em> = 2.

Then <em>L</em> - <em>R</em> = 1.

6 0
3 years ago
For a test..<br><br> Solve. 21&lt;7
alexandr1967 [171]

Answer:

21 is greater

Step-by-step explanation:

If its not.......maybe I am dumb :)

5 0
3 years ago
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