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Bess [88]
3 years ago
7

I NEED HELP ASAP! WILL GIVE BRAINLEST! Find the difference of 1/5 - (- 2 2/5)

Mathematics
2 answers:
m_a_m_a [10]3 years ago
7 0

Answer:

2.6

Step-by-step explanation:

its the answer trust me

ikadub [295]3 years ago
3 0

Answer:

2.6

Step-by-step explanation:

(1 / 5) - (-2 2/5) = 2.6

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Solve for x, the triangles are similar
elixir [45]

Answer:

x = 12

Step-by-step explanation:

Δ ESR and Δ EGF are similar, then ratios of corresponding sides are equal, so

\frac{ES}{EG} = \frac{ER}{EF} , substitute values

\frac{45}{12x-3} = \frac{55}{143} ( cross- multiply )

55(12x - 3) = 6435 ( divide both sides by 55 )

12x - 3 = 117 ( add 3 to both sides )

12x = 120 ( divide both sides by 12 )

x = 10

5 0
3 years ago
How do you write 8409.08 in words?
attashe74 [19]

Answer:

eight thousand four hundred nine and eight hundredths

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
During a game, Dennis lost 50 points in the first round and then lost 75 more points in the second round. Write an addition expr
zysi [14]

Answer:

-125

Step-by-step explanation:

add 50 and 75 =125 make in negative

5 0
3 years ago
Prove the identity.
Delvig [45]

Answer:

D. Pythagorean

Step-by-step explanation:

Given the identity

cos²x - sin²x = 2 cos²x - 1.

To show that the identity is true, we need to show that the left hand side is equal to right hand side or vice versa.

Starting from the left hand side

cos²x - sin²x ... 1

According to Pythagoras theorem, we know that x²+y² = r² in a right angled triangle. Coverting this to polar form, we have:

x = rcostheta

y = rsintheta

Substituting into the Pythagoras firnuka we have

(rcostheta)²+(rsintheta)² = r²

r²cos²theta+r²sin²theta = r²

r²(cos²theta+sin²theta) = r²

(cos²theta+sin²theta) = 1

sin²theta = 1 - cos²theta

sin²x = 1-cos²x ... 2

Substituting equation 2 into 1 we have;

= cos²x-(1-cos²x)

= cos²x-1+cos²x

= 2cos²x-1 (RHS)

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5 0
3 years ago
Hey! Can someone check my answers?
Andrew [12]
The greatest common factor is the biggest number taken from the values.

Q1. The answer is <span>A. 5y^6
</span>
20 y^{9} +5 y^{6}= 4*5 y^{9}+5 y^{6}
Since x^{a}* x^{b}  =x^{a+b}, then x^{9}= x^{3}* x^{6}

Back to our expression:
4*5 y^{9}+5 y^{6}=4*5 y^{3}*y^{6}+5 y^{6}=4 y^{3}*5y ^{6}+  5y ^{6}*1=5 y^{6} (4y ^{3} +1)
The greatest common factor is thus 5 y^{6}


Q2. The answer is <span>D. 12xy^2
</span>
12x y^{5}+60 x^{4} y^{2} -24 x^{3} y^{3}=12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}
We will use the rule  x^{a}* x^{b} =x^{a+b} to factorise the exponents:
12x y^{5}+5*12 x^{4} y^{2} -2*12 x^{3} y^{3}= \\ =12x*y^{2}*y^{3}+5*12*x* x^{3} *y^{2}-2*12x* x^{2} *y*y^{2}= \\ =12xy^{2}*y^{3}+12xy^{2}*5 x^{3} -12xy^{2}*2 x^{2} y= \\ =12xy^{2}(y^{3}+5 x^{3}-2 x^{2} y)
The greatest common factor is thus 12xy^{2}
3 0
3 years ago
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