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3 years ago

I really need help on this can someone please help me ASAP???!!!?

Mathematics

2 answers:

Charra [1.4K]3 years ago

5 0

Answer:

They all have 90 degrees angles and they have the same width and height they are just on different sides

Step-by-step explanation:

WARRIOR [948]3 years ago

3 0

Dude What Kind Of Phone You Have?

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Find the equation of the linear function represented by the table below in slope-

pav-90 [236]

Answer: $y=-3x-2$

Step-by-step explanation:

Using the values $x=-2, y=4$ and $x=1, y=-5$ , the slope is $\frac{-5-4}{1-(-2)}=-3$ .

So, the equation is:

$y-4=-3(x+2)\\\\y-4=-3x-6\\\\y=-3x-2$

3 0

1 year ago

Cheryl has some oranges. she puts the oranges into p boxes of 3 oranges each and has 2 oranges left. Find the total number of or

Umnica [9.8K]

So she had x oranges

she put the oranges into p boxes with 3 in each box and had 2 left so

total=3 times p+2 or
total=3p+2

7 0

3 years ago

Read 2 more answers

Whats the answer for number 3

Karolina [17]

cant see the full question ;/

5 0

3 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

nikdorinn [45]

Answer:

m <1 = 61

Step-by-step explanation:

We know that the angle adjacent to 130 is 50 degrees because linear angles are complementary.

Extending line t to make a triangle, we can find the angle adjacent to the angle labeled 111 ( and above it) is 69.

That makes the angle that touches line q 61 to finish the triangle

This angle and angle 1 are corresponding angles, so they are equal

<1 = 61

7 0

3 years ago

Expand the given power using the Binomial Theorem. (10k – m)5

agasfer [191]

Answer:

$(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

$(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}$

∵ The binomial is $(10k-m)^{5}$

∴ a = 10k , b = -m and n = 5

∴ $(10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}$

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ $(10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})$

∴ $(10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

5 0

3 years ago