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irina [24]
4 years ago
14

Terrell runs two timed drills at practice the first drill takes 33.5 Seconds and the second drill takes 28.2 seconds .how much t

ime does he take him to complete both drills
Mathematics
2 answers:
Advocard [28]4 years ago
7 0
61.7 seconds
33.5+28.2=61.7
Nuetrik [128]4 years ago
4 0
The answer is 61.7 seconds because if you do 33.5 seconds +20.2 seconds you look at 61.7 :-) :-)
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y= 1/2x+1

Step-by-step explanation:

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Determine the ratio of
lawyer [7]
To solve this you would divide until you can’t. I recommend staying in whole numbers:) I always divide by twos because I think it’s more efficient! 4 divided by 2 is 2 and 30 divided by 2 is 15 sooo (2 hours to 15) hope this helped!
5 0
3 years ago
The domain of both
Margarita [4]

Answer:

The domain of function h(x) is set of all real numbers.

Domain: (-∞,∞)

Step-by-step explanation:

Given:

f(x)=x-6

g(x)=x+6

the domain of both the above functions is all real number.

To find domain of :

h(x)=f(x)g(x)

Substituting functions f(x) and g(x) to find h(x)

h(x)=(x-6)(x+6)

The product can be written as difference of squares. [a^2-b^2=(a+b)(a-b)]

∴ h(x)=x^2-36

The degree of the function h(x) is 2 as the exponent of leading term x^2 is 2. Thus its a quadratic equation.

For any quadratic equation the domain is set of all real numbers.  

So Domain of h(x) is (-∞,∞)

7 0
3 years ago
11. What is the value of x in the figure at the right? (Examples 3 and 4)
natka813 [3]

Answer:

x = 11

Step-by-step explanation:

  • Vertically opposite angles are equal
  • Angles around a point add up to 360°

Therefore, 15x + 15x + 15 + 15 = 360

\implies 30x+30=360

\implies 30x=330

\implies x=11

6 0
3 years ago
If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

8 0
1 year ago
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