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finlep [7]
3 years ago
7

Combine the like terms to create an equivalent expression. \large{5t+10-4-t}5t+10−4−t5, t, plus, 10, minus, 4, minus, t

Mathematics
1 answer:
zalisa [80]3 years ago
5 0

Answer:

2(2t+3)

Step-by-step explanation:

Given the expression;

5t + 10 - 4 - t

Combining the like tems;

5t - t + 10 - 4

= (5t-t) + (10-4)

= 4t + 6

Factor out 2

= 2(2t+3)

This gives the required expression

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15 – 5x ≤ 0 and 5x + 6 ≥–14
Ad libitum [116K]
PART 1:
Let's solve your inequality step-by-step.<span><span>15−<span>5x</span></span>≤0

</span>Step 1: Simplify both sides of the inequality.<span><span><span>−<span>5x</span></span>+15</span>≤0

</span>Step 2: Subtract 15 from both sides.<span><span><span><span>−<span>5x</span></span>+15</span>−15</span>≤<span>0−15</span></span><span><span>−<span>5x</span></span>≤<span>−15

</span></span>Step 3: Divide both sides by -5.<span><span><span>−<span>5x</span></span><span>−5</span></span>≤<span><span>−15</span><span>−5</span></span></span><span>x≥3

</span>Answer:<span>x≥<span>3

PART 2:
</span></span>Let's solve your inequality step-by-step.<span><span><span>5x</span>+6</span>≥<span>−14

</span></span>Step 1: Subtract 6 from both sides.<span><span><span><span>5x</span>+6</span>−6</span>≥<span><span>−14</span>−6</span></span><span><span>5x</span>≥<span>−20

</span></span>Step 2: Divide both sides by 5.<span><span><span>5x</span>5</span>≥<span><span>−20</span>5</span></span><span>x≥<span>−4

</span></span>Answer:<span>x≥<span>−<span>4</span></span></span>
5 0
4 years ago
The equation of the circle whose center is at (4,4) and whose radius is 5 is?
nika2105 [10]
(x – h)² + (y – k)² = r²

Fill in the variables

(x – 4)² + (y – 4)² = 25
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3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

The angle 25 is the starting point and adjacent is 4 and the unknown or we will call it x is the hypotenuse.

We will use CAH

Cos25 = 4/x

We need x by itself so

x = 4/cos25

Put in calculator and put to the nearest hundredth or 2 dp

= ????

5 0
3 years ago
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sergiy2304 [10]
\textit{area of a triangle}=\cfrac{1}{2}\cdot b\cdot h\qquad &#10;\begin{cases}&#10;b=\textit{base of triangle,}\\&#10;\qquad \textit{may also be its width}\\&#10;h=\textit{height of triangle}&#10;\end{cases}&#10;\\ \quad \\&#10;b=width=18\qquad area=225&#10;\\ \quad \\&#10;225=\cfrac{1}{2}\cdot 18\cdot h\impliedby \textit{solve for "h"}
6 0
3 years ago
What’s the answer of g
Lunna [17]

Answer:

\huge\boxed{g(a-1)=\dfrac{2-a}{a}=\dfrac{2}{a}-1}

Step-by-step explanation:

g(t)=\dfrac{1-t}{1+t}\\\\g(a-1)-\text{substitute}\ t=a-1\ \text{to}\ g(t):\\\\g(a-1)=\dfrac{1-(a-1)}{1+(a-1)}=\dfrac{1-a-(-1)}{1+a+(-1)}=\dfrac{1-a+1}{1+a-1}=\dfrac{(1+1)-a}{(1-1)+a}=\dfrac{2-a}{a}

6 0
3 years ago
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