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nlexa [21]
4 years ago
7

The equation of a graph is 4x - 3y = 5. What is the x-intercept?​

Mathematics
1 answer:
Sav [38]4 years ago
6 0

Answer:(5/4,0)

Step-by-step explanation:

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PLEASE HELP ME!!! THANK YOU SO MUCH!
monitta

Answer: B / 2,6,10,14,18

Step-by-step explanation:

you were given the equation next input 1 for n.

Solve and a = 2

Then you input 2 for n. Solve and you get a =6.

Do  this up until you get to 5.

6 0
3 years ago
Monique's son just turned 2 years old and is 34 inches tall. Monique heard that the average boy will grow approximately 2 5/8 in
tatuchka [14]

Answer:

The equation representing how old Monique son is \mathbf{a = 2 + \dfrac{8}{21}(q-34)}

Step-by-step explanation:

From the given information:

A linear function can be used to represent the constant growth rate of Monique Son.

i.e.

q(t) = \hat q \times t + q_o

where;

q_o = initial height of Monique's son

\hat q = growth rate (in)

t = time

So, the average boy grows approximately 2 5/8 inches in a year.

i.e.

\hat q = 2 \dfrac{5}{8} \ in/yr

\hat q =  \dfrac{21}{8} \ in/yr

Then; from the equation q(t) = \hat q \times t + q_o

34 = \dfrac{21}{8} \times 0 + q_o

q_o = 34\  inches

The height of the son as a function of the age can now be expressed as:

q(t) = \dfrac{21}{8} \times t + 34

Then:

Making t the subject;

q - 34 = \dfrac{21}{8} \times t

t = \dfrac{8}{21}(q-34)

and the age of the son  i.e. ( a (in years)) is:

a = 2 + t

So;

\mathbf{a = 2 + \dfrac{8}{21}(q-34)}

SO;

if q (growth rate) = 50 inches tall

Then;

\mathbf{a = 2 + \dfrac{8}{21}(50-34)}

\mathbf{a = 2 + \dfrac{8}{21}(16)}

a = 2 + 6.095

a = 8.095 years

a ≅ 8 years

i.e.

Monique son will be 8 years at the time Monique is 50 inches tall.

8 0
3 years ago
Kara did a survey to determine to study the types of pets that students in her school had. She surveyed 300 students and found t
julsineya [31]

Answer:

(C)81

Step-by-step explanation:

Total Number of Students =300

Percentage who had cats, n(C) = 40%

Percentage who had both cats and dogs , n(C \cap D) = 12%

Percentage who had neither cats nor dogs, n(C \cup D)' =45%

Now:

Universal Set, U = n(C)+n(D)-n(C \cap D)+n(C \cup D)'

100 = 40 + n(D)-12+45\\n(D)=100-73\\$Percentage of those who own dogs =27\%

Therefore:

Number of Students who had dogs

=27% of 300

=81

3 0
3 years ago
In a game, Janeesa started with 0 points. She then earned 50 points, lost 80 points and earned 10 points. Which number line show
umka2103 [35]

Answer:

-20

Step-by-step explanation:

If Jeneesa started with, 0 points and she earned 50, then she would have 50. If she lost 80 points, she would go into the negatives. She would be at -30, and when she earned 10 more points, she would be at -20.

I don't know what your options look like, but look at the one(s) with -20 at the end and figure out what one best fits my explanation (and the numbers I use)!

3 0
3 years ago
Multiply the polynomials (a2 + 3a – 7) and (2a2 – a + 4). Simplify the answer.
Alex_Xolod [135]
Answer correct or incorrect

8 0
4 years ago
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