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3 years ago

PLS HELP ME ;;;-;;;

Mathematics

1 answer:

Mandarinka [93]3 years ago

5 0

Answer:

X=11

Step-by-step explanation:

9x+6= (3x+2)+(70)

Simplify: 6x+6=72

6x+6=72

6x=66

x=11

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0.10x + 185 = 0.08x + 250

geniusboy [140]

0.10x - 0.08x = 250 - 185
0.02x = 65
x = 3250

6 0

3 years ago

Consider the functions f(x) = (4/5)^x and g(x) = (4/5)^x + 6. What are the ranges of the two functions?

Nataliya [291]

Answers:
The number 0 goes in the first blank for f(x)
The number 6 goes in the second blank for g(x)

----------------------------------------------------------------------------------

Explanation:

As x gets larger and larger, the expression (4/5)^x will get smaller and smaller. What's going on is that we are repeatedly multiplying (4/5) with itself over and over if we assume x is some whole number. For example, if x = 100, then we'll have 100 copies of (4/5) multiplied out. Each time we multiply by (4/5), the result gets smaller.

The results will approach 0 but not actually get there. So this is why f(x) has the horizontal asymptote y = 0. The function values f(x) will essentially be the set of positive y values, in other words, y > 0. This covers the range for f(x). To get the range of g(x), we add 6 to the range of f(x). This is because everything on g(x) is a vertical shift upward of 6 units.

3 0

3 years ago

Please help and get points

stich3 [128]

2/4 and 1/4, 2x3=6 4x2=8, 3x1=3 3x4=12

7 0

3 years ago

Suppose Angle A is 25° and Angle B is greater than 51° but less than 57°, what are the possible measurements for Angle C?

olchik [2.2K]

The picture in the attached figure

we know that
m∠A+m∠B+m∠C=180°-----> by the sum of the internal angles of the triangle is 180 degrees
m∠A=25°

for m∠B=51°
find ∠C
∠C=180-[25+51]---------> m∠C=104°
for m∠B greater than 51°-------> m∠C must be less than 104°

for m∠B=57°
find m∠C
m∠C=180-[25+57]---------> m∠C=98°
for m∠B less than 57°-------> m∠C must be greater than 98°

therefore
98° < m∠C < 104°

4 0

3 years ago

Read 2 more answers

2. Which of the following equations are perpendicular to 2y = -3x + 1 I. II. III. y=-x-1 - 2x + 3y = -5 2x + 3y = 2 (A) I only (

Novay_Z [31]

The equation in given as ;

2y = -3x + 1

This can be written as ;

y= -3/2 x + 1/2

This means the equation has a gradient of -3/2

Let this slope , be , ---------m1

For perperdicular lines , the product of their slopes = -1 .This means if the other line has a gradient of m2 then : m1 * m2 = -1

So from the answers :

i) y= 2/3 x - 1 the slope is 2/3

m2 = 2/3

m1 * m2 = -1 -------check the if this is true by using the two values of gradient as;

-3/2 * 2/3 = - 1 ------ This is true-----equation i

II.

-2x + 3y = -5

3y = 2x -5

y= 2/3 x -5/3 -----m2 here is 2/3

m1*m2 = -1

-3/2 * 2/3 = -1 -----this is true , so ----equation ii

iii)

2x + 3y = 2

3y = -2x + 2

y= -2/3 x + 2/3 -----m2 = -2/3

m1*m2 = -1

-3/2 * -2/3 = 1 -----this is not true,,,equation iii is not perpendicular to our equation.

so, equation i and ii are perpendicular to our equation .

Answer : B i and ii only

4 0

1 year ago