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Stels [109]
3 years ago
6

Kirk and Nate are members of the same basketball team. So far this season, the team has played two games. In the first game, Nat

e scored 5 times as many points as Kirk, and in the second game, Kirk scored 3 times as many points as Nate. If Kirk and Nate have both scored a total of 42 points, which of these is a number of points that either Kirk or Nate has scored in a game this season?
Mathematics
1 answer:
Viktor [21]3 years ago
6 0
Yo is this peter cull
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Find the measure of C to the nearest degree.
Alisiya [41]

Let the ∠C be : θ

From the figure, We can see that the Side which is opposite to angle θ is measuring 7 units

Also, We can notice that Hypotenuse is 11 units

As we are dealing with opposite and hypotenuse, we can clearly use Sinθ to find out the angle θ

We know that :

\bigstar \ \ \boxed{\sf{Sin\theta = \dfrac{Opposite \ Side}{Hypotenuse}}}

\implies \sf{Sin\theta = \dfrac{7}{11}}

\implies \sf{\theta = Sin^{-1}\bigg(\dfrac{7}{11}\bigg)}

\implies \sf{\theta = 39.52^{\circ}}

<u>Answer</u> : The measure of ∠C to the nearest degree is 38°

3 0
3 years ago
Someone help me plzzz:((((
mash [69]

Step-by-step explanation:

OI= 3.2 cm (same as OH)

To find GI, use Pythagoras Theorem (a²+b²= c²)

6.8²+3.2²= 56.48

√56.48= 7.5 cm

hope it helps! please mark me brainliest

thank you! have a good day ahead

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3 0
3 years ago
Read 2 more answers
Click on the picture thing... I need help ASAP​
Aleks [24]

Answer:

A. 576 in (1728/3)

B. 1152^2 in (576x2)

C. 6912^2 in (1152x6)

D. 576^2 ft (6912/12)

6 0
3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
Which triangles are congruent by AAS?
netineya [11]
For two triangles to be congruent by AAS:
1- Two angles in the first triangle must be equal to two angles in the second triangle
2- A non included side in the first triangle is equal to a non included side in the second triangle

Now, let's check our options. We will find that:
For the two triangles UTV and ABC:
angle T = angle A
angle V = angle C
TU (non-included between angles T & V) = AB (non-included between angles A & C)

Therefore, we can conclude that:
Triangles ABC and UTV are congruent by AAS

8 0
3 years ago
Read 2 more answers
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