All categories

3 years ago

I need help with 14

Mathematics

2 answers:

In-s [12.5K]3 years ago

7 0

The answer is 1 ............................

Tomtit [17]3 years ago

4 0

His score must be 4 btw on number 15 part b. You put two ll’s and one s for Melissa while it is one l and two s

You might be interested in

Someone help quickly please!

RoseWind [281]

Answer:

translation . translation. rotation

3 0

2 years ago

V + 9 ---------- = 8 3 Can someone help me ??

bagirrra123 [75]

$\frac{v + 9}{3} = 8$

Multiply both sides by 3, because 3 is our denominator (bottom number in a fraction), and we want to get rid of the denominator.

3 × (v + 9) = 8 × 3

Simplify.

3v + 27 = 24

Then, subtract both sides by 27.

3v = 24 - 27

Simplify.

3v = -3

Divide both sides by 3.

v = -1

~Hope I helped!~

4 0

3 years ago

Ok im super confused bc like it ask me put i dont know what to put.............pls help me

Ilya [14]

Answer:

u drag the green dot thing over to the coordinate (6,8)

Step-by-step explanation:

i put a black dot where i think ur supposed to drag the green dot to

3 0

2 years ago

Answer the Blanks:

vfiekz [6]

Answer:

1. 90°

2. 90°

3. 40°

4. 45°

5. 45°

Step-by-step explanation:

Given: ΔABC,

CD⊥AB,

m∠A=50°,

m∠ACB=85°

Solution:

1. ∠ADC is a right ange, because CD⊥AB, so

$m\angle ADC=90^{\circ}$

2. ∠CDB is a right ange, because CD⊥AB, so

$m\angle CDB=90^{\circ}$

3. Consider triangle ACD. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ADC+m\angle ACD=180^{\circ}\\ \\50^{\circ}+90^{\circ}+m\angle ACD=180^{\circ}\\ \\m\angle ACD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}$

4. By Angle Addition Postulate,

$m\angle ACD+m\angle BCD=m\angle ACB\\ \\40^{\circ}+m\angle BCD=85^{\circ}\\ \\m\angle BCD=85^{\circ}-40^{\circ}=45^{\circ}$

5. Consider triangle ABC. The sum of the measures of all interior angles is always 180°, so

$m\angle A+m\angle ACB+m\angle B=180^{\circ}\\ \\50^{\circ}+85^{\circ}+m\angle B=180^{\circ}\\ \\m\angle B=180^{\circ}-50^{\circ}-85^{\circ}=45^{\circ}$