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maw [93]
2 years ago
7

Use the function a = m + 4 to find the value of a when m = 3. a =

Mathematics
2 answers:
RUDIKE [14]2 years ago
8 0

Answer:

a=m+4

a=3+4

a=7

Step-by-step explanation:

hope it helps

mark as brainlist

lora16 [44]2 years ago
8 0

Answer:

a = 7

Step-by-step explanation:

a = 3 + 4

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Can someone help me pls
Arturiano [62]

Answer:

i got u

Step-by-step explanation:

i got u

8 0
2 years ago
Sam takes a taxi in Chicago. The taxi charges $0.75 per mile and a flat fee of $1.60. If he has at most 10 dollars
Anni [7]
10 - 1.60= 8.40
8.40/0.75=11.2 miles
5 0
3 years ago
If triangle DEF is an isosceles triangle with DE is congruent to EF, find x and the measure of each side
Rudik [331]

Step-by-step explanation:

Given that,

DE = 8x - 13

EF = 5x + 17

DF = x + 21

Also,

DE = EF

which means that,

8x - 13 = 5x + 17

8x - 5x = 17 + 13

3x = 30

x = 30/3

x = 10

Now,

DE = 8x - 13 = 8×10 - 13 = 80 - 13 = 67cm

EF = 5x + 17 = 5×10 + 17 = 50 + 17 = 67cm

DF = x + 21 = 10 + 21 = 31cm

3 0
2 years ago
Jean throws a ball with an initial velocity of 64 feet per second from a height of 3 feet. Write an equation and answer the ques
Korvikt [17]

Answer:

Step-by-step explanation:

From what I understand about parabolic motion in the English system, the equation for flight is

s(t)=-16t^2+v_{0}t+h_{0}

where v_{0}t is the initial upwards velocity and

h_{0} is the initial height from which the object was launched.  Filling in that equation with those values gives you

s(t)=-16t^2+64t+3.  That's a.

In order to determine how long it will take the ball (or rocket...the problem is mixing up the 2) to reach its max height you need to put the equation into vertex form, since the vertex of a parabola is the absolute max (or min depending upon the parabola) of the function.  The absolute max is the heighest that the ball will go.  Completing the square is the way to solve this.  Begin by setting the equation equal to 0, the moving the 3 over by subtraction:

-16t^2+64t=-3

Now factor out the -16 since the leading coefficient HAS to be a positive 1:

-16(t^2-4t)=-3

Now take half the linear term (half of 4t which is 2), square it (4) and add it into the parenthesis:

-16(t^2-4t+4)=-3

BUT since you added in a 4*-16 on the left you have to add it in on the right:

-16t^2(t^2-4t+4)=-3-64

which simplifies to

-16(t-2)^2=-67

Now bring the 67 over by addition and you have your vertex:

s(t)=-16(t-2)^2+67.

The vertex is (2, 67).  The 2 stands for time, so 2 seconds, and the 67 stands for feet, so at 2 seconds the max height is 67 feet.

How long it will be in the air is found by factoring to find the zeros.  These can be found by plugging the quadratic into the quadratic formula and getting that the zeros are -0.046 and 4.046

So the quadratic starts a tiny tiny bit to the left of the origin, but for all intents and purposes we can say it starts at the origin (x = 0) and ends at

x = 4.05 seconds.  Which makes sense if you know anything about parabolic motion and physics.  The vertex indicates not only the time and the max height at that time, it also is indicative of the halfway mark.  Meaning that if it takes 2 seconds to reach its max height, it will hit the ground at 4 seconds.  And 4.05 is close enought to 4 (but since you were told to round to the nearest hundredth, that .05 matters).  Sorry it's so long, but it's not a question that can be answered with just a few sentences.

5 0
3 years ago
If a coin is tossed three times, find probability of getting
Assoli18 [71]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

‣ A coin is tossed three times.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

‣ The probability of getting,

1) Exactly 3 tails

2) At most 2 heads

3) At least 2 tails

4) Exactly 2 heads

5) Exactly 3 heads

{\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}

\star \: \tt  P(E)= {\underline{\boxed{\sf{\red{  \dfrac{ Favourable \:  outcomes }{Total \:  outcomes}  }}}}}

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

★ When three coins are tossed,

then the sample space = {HHH, HHT, THH, TTH, HTH, HTT, THT, TTT}

[here H denotes head and T denotes tail]

⇒Total number of outcomes \tt [ \: n(s) \: ] = 8

<u>1) Exactly 3 tails </u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly  \: 3 \:  tails)}  =  \red{ \dfrac{1}{8}}

<u>2) At most 2 heads</u>

[It means there can be two or one or no heads]

Here

• Favourable outcomes = {HHT, THH, HTH, TTH, HTT, THT, TTT} = 7

• Total outcomes = 8

\therefore  \sf Probability_{(at \: most  \: 2 \:  heads)}  =  \green{ \dfrac{7}{8}}

<u>3) At least 2 tails </u>

[It means there can be two or more tails]

Here

• Favourable outcomes = {TTH, TTT, HTT, THT} = 4

• Total outcomes = 8

\longrightarrow   \sf Probability_{(at \: least \: 2 \:  tails)}  =  \dfrac{4}{8}

\therefore  \sf Probability_{(at \: least \: 2 \:  tails)}  =   \orange{\dfrac{1}{2}}

<u>4) Exactly 2 heads </u>

Here

• Favourable outcomes = {HTH, THH, HHT } = 3

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 2 \:  heads)}  =  \pink{ \dfrac{3}{8}}

<u>5) Exactly 3 heads</u>

Here

• Favourable outcomes = {HHH} = 1

• Total outcomes = 8

\therefore  \sf Probability_{(exactly \: 3 \:  heads)}  =  \purple{ \dfrac{1}{8}}

\rule{280pt}{2pt}

8 0
1 year ago
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