15 POINTS!!!!!!HELP ME PLS!

The sum of two numbers is 48. If one third of one number is 5 greater than one sixth of another number, which of the following i

lubasha [3.4K]

<span>The sum of two numbers is 48.
a + b = 48
;
If one third of one number is 5 greater than one sixth of another number,
a = b + 5
multiply both sides by 6, cancel the fractions
2a = b + 30
2a - b = 30
</span><span>use elimination to solve this
a + b = 48
2a - b =30
-------------Addition eliminates b, find a
3a = 78
a =
a = 26
then
26 + b = 48
b = 48 - 26

b = 22</span>

5 0

3 years ago

According to the graph, what is the value of the constant in the equation below?

Virty [35]

According to the graph, the value of the constant in the equation below is 80 and is denoted as option D.

What is a Graph?

This is defined as a pictorial representation of data or variables in an organized manner.

From the equation: Height = constant/width

We can pick any point(4,20) on the graph.

20 = c/ 4

c = 20 × 4 = 80

Read more about Graph here brainly.com/question/19040584

#SPJ1

6 0

2 years ago

D-1/2=5/12 what is d??

EleoNora [17]

Answer:

d = $\frac{11}{12}$

Step-by-step explanation:

1) To solve for d, isolate it in the equation. So, to get rid of the $-\frac{1}{2}$ with the d, add $\frac{1}{2}$ to both sides.

$d-\frac{1}{2} = \frac{5}{12} \\d - \frac{1}{2} +\frac{1}{2} = \frac{5}{12} + \frac{6}{12} \\d = \frac{11}{12}$

Thus, d = $\frac{11}{12}$ .

6 0

3 years ago

The graph of y = x 2 has been translated 7 units to the left. The equation of the resulting parabola is _____.

strojnjashka [21]

To move the function to the left you have to increase x inside the function
think of it like g(x)=f(x+7)

the solution is
y=(x+7)^2

6 0

3 years ago

Read 2 more answers

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

2 years ago