Keelin wants to find the value of t in the equation 1/3 (15t-30) =50 Her work is shown below.

emmainna [20.7K]

9514 1404 393

Answer:

2) x = 7

3) x = 5

Step-by-step explanation:

When a transversal crosses parallel lines, all of the acute angles are congruent, and all of the obtuse angles are congruent. When it crosses at right angle, all of the angles are right angles.

2) All of the angles are right angles.

11x +13 = 90

11x = 77 . . . . . . subtract 13

x = 7 . . . . . . . divide by 11

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3) The two marked angles are acute.

16x = 80 . . . . the acute angles are congruent

x = 5 . . . . . . divide by 16

5 0

2 years ago

How many solutions does

Nikolay [14]

Answer:

A no solution

Step-by-step explanation:

y = 2x + 10

3y - 6x = 30

3(2x+10) -6x =30

6x +30 -6x =30

6x -6x = 30-30

0=0

3 0

2 years ago

If f(x) = 6x – 1 and g (x) = StartFraction x + 1 Over 6 EndFraction, which expressions can be used to verify g(x) is the inverse

tigry1 [53]

Answer:

(6 x - 1) + 1 over 6 &

6 ( x + 1 Over 6 ) - 1

Step-by-step explanation:

Edge2020

5 0

2 years ago

Read 2 more answers

Geraldine is asked to explain the limits on the range of an exponential equation using the function f(x) = 2^x. She makes these

Dafna11 [192]

Consider the exponential function $f(x)=2^x.$

By definition, the domain of a function is the set of input argument values for which the function is real and defined.

Let we take $x=2,$ then $f(2)=2^2=4$

$x=4,$ then $f(4)=2^4=16$

$x=10,$ then $f(10)=2^{10}=1024$

If we chose larger values of x, we get larger function values.

For example, If we take $f(0)=2^0=1$

$f(-3)=2^{-3}=\frac{1}{2^3}=\frac{1}{8}$

$f(-10)=2^{-10}=\frac{1}{2^10}=\frac{1}{1024}$

Thus if we choose smaller and smaller values of x. the f unction values will be smaller and smaller functions.

Thus the domain of the function is the set of all real numbers.

Thus the range is limited to the set of positive real numbers. That is, $(0,\infty)$

If we choose larger values of x, we will get larger function values, as the function values will be larger powers of 2.

If we choose smaller and smaller x values, the function values will be smaller and smaller fractions.

8 0

2 years ago

Read 2 more answers

The height of a trapezoid is 6 in. and it’s area is 72 in to the 2nd power. One base of the trapezoid is 6 inches longer than th

Vitek1552 [10]

Given:

The height of the given trapezoid = 6 in

The area of the trapezoid = 72 in²

Also given, one base of the trapezoid is 6 inches longer than the other base

To find the lengths of the bases.

Formula

The area of the trapezoid is

$A=\frac{1}{2} (b_{1} +b_{2} )h$

where, h be the height of the trapezoid

$b_{1}$ be the shorter base

$b_{2}$ be the longer base

As per the given problem,

$b_{2}=b_{1} +6$

Now,

Putting, A=72, $b_{2}=b_{1}+6$ and h=6 we get,

$\frac{1}{2} (b_{1} +b_{1}+6)(6) = 72$

or, $b_{1}+b_{1}+6 = \frac{(72)(2)}{6}$

or, $2b_{1}+6 = 24$

or, $2b_{1}=24-6$

or, $b_{1}= \frac{18}{2}$

or, $b_{1}=9$

So,

The shorter base is 9 in and the other base is = (6+9) = 15 in

Hence,

One base is 9 inches for one of the bases and 15 inches for the other base.

5 0

3 years ago