A migrating bird flies 420 miles in 12 hours . How many miles does it fly in 7 hours

Mathematics

2 answers:

____ [38]4 years ago

3 0

First find miles per hour then multiply by 7
so 420/12=35
35 times7=245 mph

belka [17]4 years ago

3 0

420 divided by 12 = 35 hours flown by the bird in 1 hour.
35 x 7(hours) = 245 miles flown in 7 hours

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Rewrite the following equation in slope-intercept form. 5x + 7y = 7 Write your answer using integers, proper fractions, and impr

pogonyaev

Answer:
y = (-5/7) x + 1

Explanation:
The slope-intercept form of the line has the following formula:
y = mx + c
where:
m is the slope
c is the y-intercept

The given is:
5x + 7y = 7

To put this in slope-intercept form, we will need to isolate the y as follows:
5x + 7y = 7
7y = -5x + 7
y = (-5/7) x + (7/7)
y = (-5/7) x + 1
where:
m is the slope = -5/7
c is the y-intercept = 1

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Find the critical numbers of the function f(x) = x6(x − 2)5.x = (smallest value)x = x = (largest value)(b) What does the Second

Marrrta [24]

Answer:

$a) x=0, x=\frac{12}{11}, x=2 \:$ b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

$f(x)=x^{6}(x-2)^{5}$

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the <em>Product Rule</em> since there is a product of two functions.

$f(x)=x^{6}(x-2)^{5}\Rightarrow f'(x)=(f*g)'(x)\\=f'g+fg'\Rightarrow (fg)'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4} \Rightarrow 6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}=0\\f'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}$

2) After that, set this an equation then find the values for x.

$x^{5}(x-2)^{4}[6(x-2)+5x]=0\Rightarrow x^{5}(x-2)^{4}[11x-12]=0\Rightarrow x_{1}=0\\(x-2)^{4}=0\Rightarrow \sqrt[4]{(x-2)}=\sqrt[4]{0}\Rightarrow x-2=0\Rightarrow x_{2}=2\\(11x-12)=0\Rightarrow x_{3}=\frac{12}{11}$

$x=0\:(smallest\:value)\:x_{3}=\frac{12}{11}\:x=2 (largest value)$

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

$f'(x)=(11x-12)(x-2)^4x^{5}$

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

$f''(x)>0 \Rightarrow Concavity\: Up\\\\f''(x)$

$f''(x)=11\left(x-2\right)^4x^5+4\left(x-2\right)^3x^5\left(11x-12\right)+5\left(x-2\right)^4x^4\left(11x-12\right)\\f''(x)=10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)$

1) Setting this to zero, as an equation:

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\\\$

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\(x-2)^{3}=0 \Rightarrow x_1=2\\x^{4}=0 \therefore x_2=0\\11x^{2}-24x+12=0 \Rightarrow x_3=\frac{12+2\sqrt{3}}{11}\:,x_4=\frac{12-2\sqrt{3}}{11}\cong 0.78$