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Schach [20]
3 years ago
10

100 POINTS!!

Mathematics
2 answers:
jeka57 [31]3 years ago
8 0

Answer:

Answer is B,C, and E

Step-by-step explanation:

I just took the Quiz

Komok [63]3 years ago
3 0

Answer:

its B,C,and E

Step-by-step explanation:

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Please don't give me a link<br><br> PLEASE HELPPP
KiRa [710]

Answer:

study more so you can understand

7 0
3 years ago
I know this isn't math but, is anyone willing to answer or help me at least answer this question?
sergeinik [125]

Answer:

What happened frnd?

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8 0
3 years ago
At the movies, a couple bought 2 medium drinks and a large popcorn for $13.85. At the same theater, a family bought 3 medium dri
yawa3891 [41]

Answer:

cost of one medium drink = $3.80

cost of one large popcorn = $6.25

Step-by-step explanation:

Let D = cost of one medium drink

Let P = cost of one large popcorn

If a couple bought 2 medium drinks and a large popcorn for $13.85:

⇒ 2D + P = 13.85

If a family bought 3 medium drinks and 2 large popcorn for a total of $23.90:

⇒ 3D + 2P = 23.90

Rewrite 2D + P = 13.85 to make P the subject:

⇒ P = 13.85 - 2D

Substitute into 3D + 2P = 23.90 and solve for D:

⇒ 3D + 2(13.85 - 2D) = 23.90

⇒ 3D + 27.7 - 4D = 23.90

⇒ D = 3.8

Substitute found value of D into P = 13.85 - 2D to find P:

⇒ P = 13.85 - (2 x 3.8) = 6.25

Therefore,

cost of one medium drink = $3.80

cost of one large popcorn = $6.25

3 0
2 years ago
What is 50 tens ×4 hundred = 40 tens × 5hundred true or false?
Vladimir79 [104]
True. 500x400=400x500
3 0
3 years ago
Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

5 0
3 years ago
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