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zimovet [89]
3 years ago
6

Please help me this is due tomorrow thank you

Mathematics
2 answers:
GaryK [48]3 years ago
4 0

Answer: D

Step-by-step explanation:

The histogram shows driving to work so A even though it looks like it may fit there is no way to assume they don’t drive at all

Hope this helps

marusya05 [52]3 years ago
3 0

Answer:

The answer is option B.

Step-by-step explanation:

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On friday, 2/5 of the students wore blue shirts and 5/12 of the students wore white shirts. What fraction of the student wore ei
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8 0
4 years ago
Find the equation of a line parallel to y=2x-4 that contains the points (5,1)
Alex73 [517]
The correct answer is
y=2x-9

Since the lines are parallel, they will gave the same slope (in this case it’s 2)

Our equation will be
y=mx+b

We know the slope is 2, and they’ve given us x and y coordinates. Plugging it in to the slope equation we get

1=2(5)+b

1=10+b

Subtract 10 from both sides and you get

-9=b

Putting it together our new parallel line will be

y = 2x - 9
5 0
3 years ago
5. Samantha wants to buy a new cell phone that costs $250. She has already
german

Answer:

A)5 is the answer because she already has 75 + 25=100 and she needs 150 more so 150÷25=5

6 0
2 years ago
A number is chosen at random from 1 to 25 find the probability of selecting a number less than 7
Anna11 [10]

Answer:

6/25

Step-by-step explanation:

6 numbers (from 1 - 25) are less than 7.  There are 25 numbers in total.  This makes the probability 6/25.

8 0
4 years ago
Given: C∉ BD ,△ABC D∈ ray BC ,AB=AC=BC Prove: BD>DA>AB
jonny [76]

Triangle ABC is equilateral, because AB=BC=AC=a. Then

m∠A=m∠B=m∠C=60°.

Let point D lie on the ray BC to the right from points B and C and let CD=x. Then BD=a+x, AB=a.

Consider triangle ACD. In this triangle, m∠ACD=180°-m∠ACB=180°-60°=120°.

By the cosine theorem,

AD^2=AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos \angle ACD,\\\\AD^2=a^2+x^2-2\cdot a\cdot x\cdot \cos 120^{\circ},\\\\AD^2=a^2+x^2+ax,\\\\AD=\sqrt{a^2+x^2+ax}.

Since a^2+x^2+ax=a^2+x^2+ax+ax-ax=(a+x)^2-ax, then

AD^2=(a+x)^2-ax

and

AD^2=a^2+x^2+ax>a^2=AB^2\Rightarrow AD>AB.

Therefore, you get double inequality

AB or BD>AD>AB.

3 0
3 years ago
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