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choli [55]
3 years ago
10

Write a function rule for the table days/cost to rent a truck

Mathematics
2 answers:
aivan3 [116]3 years ago
8 0
Its c hope that helps text me back if you want to know why it is.
kirill115 [55]3 years ago
5 0

c is your answer

hope this helps

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What is the diameter of a circle whose radius is 6.45 meters
Vikentia [17]
12.9
Diameter is just the radius times 2
6 0
3 years ago
Try this B E Given: BELO, BOLE Prove ABEL ALOB Complete the Proof Statements 1. 2. BO=LE 3. 4. ∆BEL =∆LOB REASONS 1.GIVEN 2. 3.
AnnZ [28]

The given congruency of the sides \mathbf{\overline{BE}} and \mathbf{\overline{LO}}  and \mathbf{\overline{BO}} and \mathbf{\overline{LE}} as well

as the congruency of the common side \mathbf{\overline{BL}} gives.

ΔBEL ≅ ΔLOB by SSS congruency postulate

<h3>Which values correctly completes the table?</h3>

The completed two column proof is presented as follows;

Statement {}                                          Reasons

1. \underline{\overline{BE} \cong \overline{LO}}                       {}                  1. Given

2. \overline{BO} ≅ \overline{LE}                       {}                 2.

3. \underline{\overline{BL} \cong \overline{BL}}                       {}                  3. <u>Reflexive property of congruency</u>

4. ΔBEL ≅ ΔLOB                     {}           4. <u>SSS congruency postulate</u>

Side-Side-Side, SSS, congruency postulate states that if three sides of

one triangle are congruent to three sides of another triangle, the two

triangles are congruent.

Learn more about different congruency postulates here:

brainly.com/question/1495556

7 0
2 years ago
Find the area of the regular pentagon.
IceJOKER [234]

Answer: 172.05

Hope this helps

5 0
3 years ago
Find the zeros (x-3)(x+5)=0
maxonik [38]
The product is equal zero if one of the factors is equal zero.Therefore:
(x-3)(x+5)=0\iff x-3=0\ \vee\ x+5=0\\\\\boxed{x=3\ \vee\ x=-5}

3 0
3 years ago
What is the value of (7 + 2i)(3 - i)?
Harrizon [31]
There are two ways to work this out: normal variables or using "imaginary" numbers.

Normal variables:
(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}

Imaginary numbers:
Using the result from earlier:
21-i-2i^{2}
Now since i = \sqrt{-1}, then the expression becomes:
21-i-2i^{2}\\21-i-2(\sqrt{-1})^{2}\\21-i+(-2)(-1)\\21-i+2\\\\23-i
7 0
3 years ago
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