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Anton [14]
3 years ago
10

Describe two methods that can be used to find the area of the composite figure.

Mathematics
2 answers:
sammy [17]3 years ago
7 0

<u><em>Sample Response:</em></u>

Find the area of the larger rectangle, then subtract the area of the two smaller rectangles in the corners. Separate the figure into two or three rectangles, and add the areas.

12345 [234]3 years ago
5 0
Multiplication and PxW
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Which of the following is/are on a credit report ?
Liono4ka [1.6K]

Answer:

companies who have checked persons credit.

Step-by-step explanation:

the more companies you allow to check credit, could be an indication of applying for multiple credit lines, which is a negative.

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4 years ago
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Which expression is equivalent to cot t sec t?
Solnce55 [7]

Answer: csc(t)

Step-by-step explanation:

Alright, lets get started.

The given expression is given as :

cot( t) \times sec (t)

We know quotient identity as :

cot(t)=\frac{cos(t)}{sin(t)}

Similarly, we know reciprocal identity as :

sec(t)=\frac{1}{cos(t)}

lets plug the value of cot and sec in given expression

\frac{cos(t)}{sin(t)}  \times \frac{1}{cos(t)}

cos will be cancelled, remaining will be

\frac{1}{sin(t)}

Using reciprocal identity again, that will equal to :

csc(t)  ................... Answer (A)

8 0
3 years ago
QUESTION 1 - 1 POINT
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Answer + Step-by-step explanation:

the inequality by finding the boundary line, then shading the appropriate area.

y > − 5/2x − 5

(anything in shaded region)

8 0
2 years ago
Kevin is 6 years older than Timothy. 4 years ago, Kevin was twice as old as Timothy. Find their present ages.
Novosadov [1.4K]

Step-by-step explanation:

Hope it helps.

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6 0
3 years ago
Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose tha
san4es73 [151]

Answer:

0.2364

Step-by-step explanation:

We will take

Lyme = L

HGE = H

P(L) = 16% = 0.16

P(H) = 10% = 0.10

P(L ∩ H) = 0.10 x p(L U H)

Using the addition theorem

P(L U H) = p(L) + P(H) - P(L ∩ H)

P(L U H) = 0.16 + 0.10 - 0.10 * p(L u H)

P(L U H) = 0.26 - 0.10p(L u H)

We collect like terms

P(L U H) + 0.10P(L U H) = 0.26

This can be rewritten as:

P(L U H)[1 +0.1] = 0.26

Then we have,

1.1p(L U H) = 0.26

We divide through by 1.1

P(L U H) = 0.26/1.1

= 0.2364

Therefore

P(L ∩ H) = 0.10 x 0.2364

The probability of tick also carrying lyme disease

P(L|H) = p(L ∩ H)/P(H)

= 0.1x0.2364/0.1

= 0.2364

3 0
3 years ago
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