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labwork [276]
3 years ago
7

A survey asked a group of students to list their eye color. The results of the survey are shown in the graph.

Mathematics
2 answers:
Evgesh-ka [11]3 years ago
7 0

Answer:

b?

Step-by-step explanation:

drek231 [11]3 years ago
3 0
I think the answer is b
You might be interested in
Find cosØ if sinØ=-12/13 and tanØ&gt;0.<br><br> A) -5/12<br> B) -5/13<br> C) 12/5<br> D) -13/12
mestny [16]

Answer:

-5/13

Step-by-step explanation:

sin theta = opp / hyp

sin theta = -12 /13

we can find the adj side by using the pythagorean theorem

adj^2 + opp ^2 = hyp^2

adj^2 +(-12)^2 = 13^2

adj^2 +144 =169

adj^2 = 169-144

adj^2 = 25

Taking the square root of each side

adj = ±5

We know that it has to be negative since it is in the third quad

adj = -5

cos theta = adj / hyp

cos theta = -5/13

8 0
3 years ago
Read 2 more answers
What is the equation of the line perpendicular to y=2x+3.14 going through the points (2,4)?
inessss [21]

Answer:

x+2y-10=0

Step-by-step explanation:

Consider an equation: y=mx+c

Here, m is the slope of the line and c is the y-intercept.

Given equation is y=2x+3.14

Here, slope is m=2

As product of slopes of two perpendicular lines is equal to -1, slope of the required line is m_1=\frac{-1}{m}=\frac{-1}{2}.

Let (x_1,y_1)=(2,4)

Equation of the required line is y-y_1=m_1(x-x_1)

y-4=\frac{-1}{2}(x-2)\\2(y-4)=-1(x-2)\\2y-8=-x+2\\x+2y-8-2=0\\x+2y-10=0

8 0
3 years ago
What is the range of the exponential function f(x)=2•7x
sergeinik [125]

f(x) cannot be 0 or less than zero, but can approach+ infinity

Answer is (0, ∞)

5 0
3 years ago
Now there is a square city of unknown size with a gate at the center of each side. There is a tree 20 b from the north gate. Tha
Brums [2.3K]

Answer:

The length of each side of the city is 250b

Step-by-step explanation:

Given

a = 20 --- tree distance from north gate

b =14 --- movement from south gate

c = 1775 --- movement in west direction from (b)

See attachment for illustration

Required

Find x

To do this, we have:

\triangle ADE \sim \triangle ACB --- similar triangles

So, we have the following equivalent ratios

AE:DE = AB:CB

Where:

AE = 20\\ DE = x/2 \\ AB = 20 + x + 14 \\ CB = 1775

Substitute these in the above equation

20:x/2 = 20 + x + 14: 1775

20:x/2 = x + 34: 1775

Express as fraction

\frac{20}{x/2} = \frac{x + 34}{1775}

\frac{40}{x} = \frac{x + 34}{1775}

Cross multiply

x *(x + 34) = 1775 * 40

Open bracket

x^2 + 34x = 71000

Rewrite as:

x^2 + 34x - 71000 = 0

Expand

x^2 + 284x -250x - 71000 = 0

Factorize

x(x + 284) -250(x + 284)= 0

Factor out x + 284

(x - 250)(x + 284)= 0

Split

x - 250 = 0 \ or\ x + 284= 0

Solve for x

x = 250 \ or\ x =- 284

x can't be negative;

So:

x = 250

6 0
3 years ago
NO LINKS
Elza [17]
So the answer is 30 packages but I did 25 times 20 = 500 and then 25 times 10 = 250
So. 30 = 750
6 0
2 years ago
Read 2 more answers
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