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2 years ago

The number 68 is increased to 73. What is the percentage by which the number was

Mathematics

1 answer:

Murljashka [212]2 years ago

3 0

Answer: 1.1%

Step-by-step explanation:

To get a percentage you want to divide the ending number by the base number

73 / 68 = 1.07

It says to round to the neartest tenth, when rounding if the number is 5 or above you round up, if under 5 your number stays the same

7 is higher than 5 so we round up

1.1%

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Need help with this math problem

zheka24 [161]

Answer:

m<1 = 51

m<2 = 61

m<3 = 29

Step-by-step explanation:

three angles of a triangle = 180

so knowing that add 68 + 61 to get 129

now subtract cause the last angle will = 180-129 = 51

for angle two is 61 cause their congruent

now we have one angle = 90 and another 61

add them 90+61= 151

now subtract 180-151 = 29

6 0

2 years ago

If f(x)=4x^2 and g(x)=x+1, find (fog)(x)

iren [92.7K]

Answer: The value of $(f_\circ g)(x)$ is $4(x+1)^2$ .

Step-by-step explanation:

Given: $f(x) = 4x^2 \text { and } g(x) = x+1$

To find: $(f_\circ g)(x)$

As we know it is composition function which means that g(x) function is in f(x) function.

So we have

$(f_\circ g) (x) = f[g(x)]$

$\Rightarrow( f_\circ g)(x)= f(g(x)) = 4(g(x))^2$

Now substitute the value of g(x) we get

$(f_\circ g)(x)= 4(x+1)^2$

Hence, the value of $(f_\circ g)(x)$ is $4(x+1)^2$ .

5 0

3 years ago

If y varies directly as the square root of x and y=12 when x = 16 , find x when y = 15

a_sh-v [17]

= 19 this is the answer I'm guessing since 16-12= 4 so 16 is 4 more than 12

3 0

3 years ago

a sequence has a first number of 3 and a half common difference of 4 what is the recursive rule of nth term

ELEN [110]

Answer:

$f(n) = f(n-1) + 8$ for $n > 1$

Step-by-step explanation:

Given

$f(1) = 3$ -- First Term

$\frac{1}{2}d = 4$ --- half common difference

Required

Find the recursive rule

First, we calculate the common difference

$\frac{1}{2}d = 4$

Multiply through by 2

$2 * \frac{1}{2}d = 2 * 4$

$d = 8$

The second term of the sequence is:

$f(2) = 3 + 8 = 11$

The third term is:

$f(3) = 11 + 8 = 20$

So, we have:

$f(1) = 3$

$f(2) = 3 + 8$

Substitute f(1) for 3

$f(2) = f(1) + 8$

Express 1 as 2 - 1

$f(2) = f(2-1) + 8$

Substitute n for 2

$f(n) = f(n-1) + 8$

Similarly:

$f(3) = 11 + 8$

Substitute f(2) for 11

$f(3) = f(2) + 8$

Express 2 as 3 - 1

$f(3) = f(3-1) + 8$

Substitute n for 3

$f(n) = f(n-1) + 8$

Hence, the recursive is:

$f(n) = f(n-1) + 8$ for $n > 1$

8 0

2 years ago

Solve the following quadratic equation for all values of x in simplest form.

denis-greek [22]

Answer:

$x=\sqrt{7},\:x=-\sqrt{7}$

Step-by-step explanation:

Start by rewriting the equation to standard form $ax^2+bx+c=0$ .

$5(x^2-10)+5=-10,\\5x^2-50+5=-10,\\5x^2-35=0$ .

Factor out the 5, then isolate $x^2$ :

$5x^2-35=0,\\5(x^2-7)=0,\\x^2-7=0,\\x^2=7,\\x=\pm \sqrt{7}.$ .

Therefore, the solutions to this quadratic are:

$x=\sqrt{7},\:x=-\sqrt{7}$

8 0

3 years ago