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2 years ago

Given h(a) = 5x – 2. find h(5)

Mathematics

2 answers:

Maru [420]2 years ago

8 0

Answer:

Step-by-step explanation:

h(a) = 5x-2

h(5) = 5(5)-2

h(5) = 25-2

h(5) = 23

mafiozo [28]2 years ago

8 0

Answer:

Step-by-step explanation:

Plug in 5 as x

h(x) = 5x - 2

h(5) = 5(5) - 2

h(5) = 25 - 2

h(5) = 23

So, h(5) = 23

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3 years ago

If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

4 0

2 years ago

Read 2 more answers

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This year is yer number 2 , so t = 2
which means that the next year will be t = 3

So, Epsilon-delta should order :
42000 (1.3 ^ (3/2))
= 62,254 Tags

Hope this helps

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2 years ago

Given h(a) = 5x – 2. find h(5)​

Given h(a) = 5x – 2. find h(5)