All categories

3 years ago

If f (X) = 4x + 18, find the value of x so that f (X) = 34.

Mathematics

2 answers:

rjkz [21]3 years ago

4 0

Answer:

solution

f(x)=4x+18

f(x)=34

now,

34=4x+18

or,34-18=4x

or,16/4=x

or,x=4

hence,x=4

Rudiy273 years ago

3 0

X=4 because 34-18 equals 16 and 4(4) equals 16

You might be interested in

The total number of students enrolled in MATH 123 this semester

Kay [80]

Answer:

5796 people

Step-by-step explanation:

.28 percent of 5780 is 16.184 so added 5,780+16.184=5,796.184 but rounded to a whole person is 5,796!

7 0

3 years ago

Jos donut express earned 4,378 at a festival by selling chocolate or vanilla donuts for 2 dollars each if they sold 978 chocolat

ANEK [815]

2422 because 978x2=1956
4378-1956=2422

7 0

3 years ago

Read 2 more answers

1<br> Solve:<br> -8–2k=-2(k+4)

Olegator [25]

Answer:

zero=0

Explanation:

-8-2k=-2(k+4)

-8-2k=-2k-8

-2k+2k=-8+8

zero=0

4 0

3 years ago

5x - 2y = 18<br> -5x + 3y = -22

jeka94

Add the two equations
since the 5x and -5x cancel out
you get y=-4

plug y=-4 into the first equation
solve for x
x=2

THE ANSWER IS: (2,-4)

8 0

3 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago