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3 years ago

5h-4h^3-2 I need help

Mathematics

1 answer:

natima [27]3 years ago

8 0

Answer:

Step-by-step explanation:

What's the complete question?

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The United States Congress contains 435 representatives and 100 senators. You want to make groups that each have the same number

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

The greatest common factor of 435 and 100 is 5, so if you make 5 groups, each group will have 87 representatives and 20 senators.

7 0

3 years ago

jasmine is 5 years younge router than twice her sister's age. If Jasmine sister is 22 years old then how old is jasmine

Margarita [4]

I think she is 39 because twice jasmine age 22 * 2 is 44 minus 5 which is 39

6 0

3 years ago

Read 2 more answers

A trapezoid has 8.2 millimeters base and 5.2 millimeters base, as well as 10 millimeters height. What is the area?

Savatey [412]

Answer:

67 mm

Step-by-step explanation:

use area of a trap formula:

b1+b2 /2*h

b1=base

b2=base

h=height

in our case it was, 8.2+5.2/2*10

6 0

3 years ago

In his suitcase, Jack has 3 shirts, 4 pants, 2 socks (pairs of socks), and 2 shoes (pairs of shoes). How many unique ways can Ja

kipiarov [429]

Answer:

Step-by-step explanation:

the number of shirts are 3, number of pants 4, number of socks pairs 2 and 2 pairs of shoes.

first lets start with shirts and pants, number of unique combinations are,

(for each shirt there are 4 different pant combinations) = $(3)(4)$

=12

similiarly for each shirt pant pair there are 2 shoes and 2 socks pairs.

thus, total number of combinations are=

$(12)(2)(2)$

= 48

8 0

4 years ago

Given: C∉ BD ,△ABC D∈ ray BC ,AB=AC=BC Prove: BD>DA>AB

jonny [76]

Triangle ABC is equilateral, because AB=BC=AC=a. Then

m∠A=m∠B=m∠C=60°.

Let point D lie on the ray BC to the right from points B and C and let CD=x. Then BD=a+x, AB=a.

Consider triangle ACD. In this triangle, m∠ACD=180°-m∠ACB=180°-60°=120°.

By the cosine theorem,

$AD^2=AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos \angle ACD,\\\\AD^2=a^2+x^2-2\cdot a\cdot x\cdot \cos 120^{\circ},\\\\AD^2=a^2+x^2+ax,\\\\AD=\sqrt{a^2+x^2+ax}.$

Since $a^2+x^2+ax=a^2+x^2+ax+ax-ax=(a+x)^2-ax,$ then

$AD^2=(a+x)^2-ax$

and

$AD^2=a^2+x^2+ax>a^2=AB^2\Rightarrow AD>AB.$

Therefore, you get double inequality

$AB$ or $BD>AD>AB.$

3 0

3 years ago