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anygoal [31]
2 years ago
6

Please help me I’ll give brainliest

Mathematics
2 answers:
umka21 [38]2 years ago
7 0

Answer:

Rational

Step-by-step explanation:

2.4 is a terminating decimal

beks73 [17]2 years ago
5 0

Answer:

rational

Step-by-step explanation:

since rational numbers include both negative,positive , positive decimals negative and positive decimals, fractions..

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n the Question ax + bx, a is a decimal and bis a fraction. How do you decide whether to write a as a fraction or b as a decimal?
ASHA 777 [7]
It’s a number hjcjckchcjghh Igh do
3 0
3 years ago
Write expanded form for six hundred and four hundred thirteen thousandths
lakkis [162]
64,013,000 =
<span>six hundred and four hundred thirteen thousandths =
60,000,000 + 4,000,000+10,000+3,000</span>
4 0
3 years ago
Triangle ABC is graphed below with vertices at A (-4,5), B (-6, -4) and C (4,-4). Use the following rule to produce
guapka [62]

Step-by-step explanation:

A(-4, 5) => A'(-4+2, -5) = A'(-2, -5)

B(-6, -4) => B'(-6+2, -(-4)) = B'(-4, 4)

C(4, -4) => C'(4+2, -(-4)) = C'(6, 4)

Hence the new coordinates are A'(-2, -5), B'(-4, 4) and C'(6, 4).

7 0
3 years ago
Select all the correct locations on the table.
AlladinOne [14]
Average rate of change: r=[f(b)-f(a)]/(b-a)

r=-60→[f(b)-f(a)]/(b-a)=-60

b=5; f(b)=-213; a=1; f(a)=27
(-213-27)/(5-1)=(-240)/4=-60

Answer: The <span>two points in the table which create an interval with an average rate of change of -60 are:
x     f(x)
1      27
5   -213</span>
7 0
3 years ago
Read 2 more answers
The angle θ1\theta_1 θ 1 ​ theta, start subscript, 1, end subscript is located in Quadrant II\text{II} II start text, I, I, end
melomori [17]

Answer:

\dfrac{3\sqrt{13}}{11}

Step-by-step explanation:

Given that the angle \theta_1  is located in Quadrant II; and

\cos(\theta_1)=-\dfrac{2}{11}

In Quadrant II, x is negative and y is positive.

\cos(\theta)=\dfrac{Adjacent}{Hypotenuse},\sin(\theta)=\dfrac{Opposite}{Hypotenuse}\\$Adjacent=-2\\Hypotenuse=11\\

To find \sin(\theta_1), we first determine the opposite angle of \theta_1.

This will be done using the Pythagoras theorem.

Hypotenuse^2=Opposite^2+Adjacent^2\\11^2=Opposite^2+(-2)^2\\Opposite^2=121-4=117\\Opposite=\sqrt{117}=3\sqrt{13}

Therefore:

\sin(\theta_1)=\dfrac{Opposite}{Hypotenuse}=\dfrac{3\sqrt{13}}{11}

6 0
2 years ago
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