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3 years ago

SOMEBODY PLEASE HELP

Mathematics

2 answers:

Ivanshal [37]3 years ago

7 0

AB---------DE

BC---------EF

AC----------DF

uranmaximum [27]3 years ago

4 0

AB corresponds to DE
BC corresponds to EF
AC corresponds to DF

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1/2 + 1 1/4 less than 2

Wittaler [7]

Answer:

No, it only equals 1 3/4.

Step-by-step explanation:

You first need to break up 1/2. You should know it equals 2/4, so adding 1 1/4 and 2/4, it will equal 1 3/4

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3 years ago

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NEED HELP!! Trigonometry questions

anzhelika [568]

Answer:

Step-by-step explanation:

Question 1: Assumption: This is a 30-60-90 triangle.

Remember that the sides of a 30-60-90 triangle are in the ratio 1:√3:2

The side opposite the 90° angle is 16, so the side opposite the 30° angle is 16/2 = 8

x = 8 units.

:::::

Question 2: Assumption: This is an isosceles triangle.

Draw the altitude to the vertex angle and you get a 30-60-90 triangle.

The side opposite the 90° angle has length 22, so the side opposite the 30° angle has length 11.

x/2 = 11

x = 22 units

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Question 3: Assumption: This is a 45-45-90 triangle.

The sides of a 45-45-90 triangle are in the ratio 1:1:√2

The sides opposite the 45° angles are 19 and x.

x = 19

:::::

Question 4: Assumption: This is an isosceles triangle.

x = 13 units

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Question 5: Assumption: This is a right triangle.

sin(54°) = x/45

x = 45sin(54°) ≅ 36.4 units

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Question 6: Assumption: This is a right triangle.

sin(35°) = z/23

z = 23sin(35°) ≅ 13.2 units

4 0

3 years ago

One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number o

Zielflug [23.3K]

To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

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3 years ago

What are two ratio equivalent to 5/9

ZanzabumX [31]

10/18 and 15/27 These 2 fractions here could be possible equivalent fractions.

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4 years ago

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X ^ 2 + y ^ 2 + 6y - 67 = 2y - 6x; circumference

garri49 [273]

Answer:

(x+3)^2+(y+2)^2=80