3 years ago

Can someone please help me./ solve.

Mathematics

2 answers:

Zina [86]3 years ago

7 0

Answer:f=99

Step-by-step explanation

solve for f by simplifying both sides of the equation, then isolating the variable.

Pavel [41]3 years ago

3 0

Answer:

F is 99

Step-by-step explanation:

99 ÷ 77 = 1 2/7

1 2/7 - 4/7 = 5/7

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When f(x)= -3,what is x

White raven [17]

F(x) = -3.

x = f⁻¹(-3)

x = f inverse of (-3).

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3 years ago

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goblinko [34]

D. 96°
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3 years ago

Quien me ayuda? pls!!!

posledela

Answer:

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Step-by-step explanation:

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3 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

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Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago

Target sells pencils at 5 for $1.25. <br> how many pencils could you but for $22?

amm1812

22 divided by 1.25 = 17.5
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6 0

3 years ago

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