Gcf of 28 and84 and 65 39 and 75 and 45

1. Remember what we know about vertical angles and solve for x. (SHOW WORK)

EastWind [94]

Answer:

Ans.1.) X = 7.

Ans.2.a.) AC =JL.

Ans.2.b.) BC = KL.

Step-by-step explanation:

For Question 1.

As we know vertically opposite angles are equal.

$\therefore (x+16)\° = (4x-5)\°\\\therefore (5+16)\° = (4x-x)\°\\ \therefore (3x) = (21)\\\therefore x = 7\\$

For Question 2.a)

$In\ \triangle ABC and \triangle JKL \\AB \cong JK\ \textrm{Given}\\\angle CAB \cong \angle LJK\ \textrm{each measure angle of 90\°}\\AC \cong JL\ \textrm{this is the required information to prove the triangles are congruent by SAS postulate}\\\therefore \triangle ABC \cong \triangle JKL\ \textrm{ By SAS postulate}$

For Question 2.b)

$In\ \triangle ABC and \triangle JKL \\AB \cong JK\ \textrm{Given}\\\angle CAB \cong \angle LJK\ \textrm{each measure angle of 90\°}\\BC \cong KL\ \textrm{this is the required information to prove the triangles are congruent by HL Theorem}\\\therefore \triangle ABC \cong \triangle JKL\ \textrm{ By Hypotenuse length Theorem}$

3 0

2 years ago

Hayden likes building radio controlled sailboats with her father. One of the sails shaped like a tight triangle has side lengths

disa [49]

Answer:

the scaled sail has lengths 1.5 inches, 2 inches, and 2.5 inches. the diagram is attached.

Step-by-step explanation:

a scale factor of 1/4 means that the lengths of each side of the sail will be reduced by 0ne-forth (0.6) its original length. Therefore the lengths of the scale drawing before and after scaling are:

6 inches = 6 × 1/4 = 6 × 0.25 = 1.5 inches

8 inches = 7 × 0.25 = 2 inches

10 inches = 10 × 0.25 = 2.5 inches.

Therefore the scaled sail has lengths 1.5 inches, 2 inches, and 2.5 inches as shown in the diagram attached.

6 0

3 years ago

Circle O has a circumference of approximately 44 in. What is the approximate length of the diameter, d? 7 in. 14 in. 22 in. 44 i

katrin [286]

Circumference = 3.1415 (pi) x diameter
44 = 3.1414 x diameter

Diameter = 14.006 or 14 (aprox)

6 0

3 years ago

Read 2 more answers

What numbers are a distance of 4\6 unit from 3\6 on a number line?<br><br>Please help me out.

sweet [91]

The numbers that are a distance of 4/6 unit from 3/6 on a number line are -1/6 and 7/6

<h3>What is the distance on a Number Line?</h3>

We want to know what numbers are a distance of 4/6 unit from 3/6 on a number line.

Add 4/6 to 3/6, and also subtract 4/6 from 3/6. That will give the two numbers.

3/6 + 4/6 = 3/6 + 4/6 = 7/6

3/6 - 4/6 = -1/6

Thus, the numbers that are a distance of 4/6 unit from 3/6 on a number line are -1/6 and 7/6

Read more about Distance on a number Line at; brainly.com/question/17143316

#SPJ1

3 0

1 year ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = 3 (*sqaure root sign*) t (20 − t), [0,

mafiozo [28]

9514 1404 393

Answer:

maximum: 15∛5 ≈ 25.6496392002
minimum: 0

Step-by-step explanation:

The minimum will be found at the ends of the interval, where f(t) = 0.

The maximum is found in the middle of the interval, where f'(t) = 0.

$f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)$

This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...

f(5) = (∛5)(20-5) = 15∛5

The absolute maximum on the interval is 15∛5 at t=5.

5 0

2 years ago