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3 years ago

10

Only 3% of turtle hatchlings will live to become breeding adults. How many turtles must have been born if the current number of

breeding adults is 1,200?

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

116400 I believe not 100% sure tho

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Slav-nsk [51]

The answer is a, because the absolute value of x(the actual weight of the product) minus the target weight of 18.25 needs to be greater that the target weight difference of .36 so that it is outside the range. So basically, any number outside of the range will be correct.You can plug in number that you know are in/ out of the acceptable range of numbers for this problem in.
|18.62-18.25|>.36
.37>.36
18.62 is not in range.

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3 years ago

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What the answer pls…

Valentin [98]

The slope is a positive and constant increase of 1 over 2 and the y - intercept is located at 8 due to the existing amount of snow on the ground. hope this helped you might want to word it a little differently.

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3 years ago

Find the missing measure if a and b are the legs of the right triangle and c is the hypotenuse, with a = 11 and c =18.

never [62]

Answer:

$\sqrt{203}$

Step-by-step explanation:

1. $11^{2} + b^{2} = 18^{2}$

2. $b^{2} =203$

3. b = $\sqrt{203}$

Can't be simplified.

3 0

3 years ago

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Solve the equation<br> -5=y for x.

sesenic [268]

Answer: 5 = y

Step-by-step explanation:

7 0

3 years ago

Please answer question three and four if you can :)<br> Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

4 years ago