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3 years ago

I need help as quickly as possible please Necesito ayuda lo más rápido posible por favor

Mathematics

2 answers:

kenny6666 [7]3 years ago

8 0

1
-1
-3
3

Here’s the answer ^^

AVprozaik [17]3 years ago

7 0

Answer:

1 | 1

0 | -1

-1 | -3

2 | 3

Step-by-step explanation:

To fill the table, simply replace the x in the function with the value given.

For instance,

f(1) = 2(1) - 1 = 1

f(0) = 2(0) -1 = -1

f(-1) = 2(-1) - 1 = -2 - 1 = -3

f(2) = 2(2) - 1 = 4 - 1= 3

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Writing a Two-Column Proof

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For the blank on the left, the answer is "angle ABC" without quotes of course. This is because we're using the substitution property which is the reasoning for statement 4. To elaborate, we connect statement 1 and statement 3 to get to statement 4.

The last statement, statement 5, is simply a repeat of the claim we want to prove. So you'll just copy/paste where it says "prove" as your last statement. Always start with the "given" and end up with "prove" is how I think of it.

The reasoning for the last statement is "definition of congruence". If two angles are congruent, then they have the same angle measure.

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3 years ago

Amber saved a total of $3.20over 5 weeks .She saved the same amount of money each week.How much money did amber save each week

Anastasy [175]

$0.64

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4 years ago

A. 90 B. 70 C. 55 D. 35

schepotkina [342]

Answer:

A. 90

Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve for r. Give an exact answer. 1/2r-3=2(4-3/2r)

Marysya12 [62]

We can use the distribution method for this:

1/2r - 3 = 2(4 - 3/2r)

Multiply 2 by everything inside the parentheses:

1/2r - 3 = 8 - 3r.

Move variables to one side, and numbers to another:

1/2r + 3r = 8 + 3.

3 1/2r = 11.

7/2r = 11.

Divide by 7/2:

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4 years ago

Will give brainiest In △ABC, angle bisectors

EleoNora [17]

Answer:

$m\angle CAB=102\dfrac{6}{7}^{\circ}\\ \\m\angle ABC=51\dfrac{3}{7}^{\circ}\\ \\m\angle ACB=25\dfrac{5}{7}^{\circ}$

Step-by-step explanation:

AK is angle A bisector, then

$m\angle BAK=m\angle KAC=x^{\circ}$

BL is angle B bisector, then

$m\angle ABL=m\angle CBL=y^{\circ}$

Consider triangle ABL. The sum of the measures of all interior angles in this triangle is $180^{\circ},$ then

$m\angle BAL+m\angle ALB+m\angle LBA=180^{\circ}\\ \\2x+2y+y=180\\ \\2x+3y=180$

Consider triangle ABK. In this triangle,

$m\angle AKB=180^{\circ}-2x^{\circ} \ [\text{Supplementary angles}]$

The sum of the measures of all interior angles in this triangle is $180^{\circ},$ then

$m\angle BAK+m\angle AKB+m\angle KBA=180^{\circ}\\ \\x+(180-2x)+2y=180\\ \\2y-x=0$

Hence,

$x=2y\\ \\2(2y)+3y=180\\ \\4y+3y=180\\ \\7y=180\\ \\y=\dfrac{180}{7}=25\dfrac{5}{7}\\ \\x=\dfrac{360}{7}=51\dfrac{3}{7}$

Find the measures of the triangle ABC:

$m\angle CAB=2x^{\circ}=102\dfrac{6}{7}^{\circ}\\ \\m\angle ABC=2y^{\circ}=51\dfrac{3}{7}^{\circ}\\ \\m\angle ACB=180^{\circ}-102\dfrac{6}{7}^{\circ}-51\dfrac{3}{7}^{\circ}=25\dfrac{5}{7}^{\circ}$

8 0

4 years ago