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3 years ago

15 POINTS

2 answers:

8 0

So m = what he runs during the week
He normally runs 10 blocks on the weekend, but he ran 12 more blocks this weekend so the answer is ... D m+22

trasher [3.6K]3 years ago

4 0

Answer:

D m+22

Step-by-step explanation:

You might be interested in

Find the product and express your answer in scientific notation. (1.3 x 10-6)(4 x 1010)

sveticcg [70]

$(1.3\cdot10^{-6})(4\cdot10^{10})=(1.3\cdot4)(10^{-6}\cdot10^{10})=5.2\cdot10^{-6+10}\\\\=\boxed{5.2\cdot10^4}$

3 0

3 years ago

If a quadratic function has x-intercepts at (-7,0) and 10,0), where would

madam [21]

Answer:

x = 1.5

Step-by-step explanation:

Since the axis of symmetry is between the 2 x intercepts, we can find it by finding the average of the x-coordinates of the 2 points.

(-7 + 10) / 2

3/2

= 1.5

So, the axis of symmetry is at x = 1.5

8 0

2 years ago

What is the domain and range?

ycow [4]

Answer:

Domain- the set of possible values of the independent variable or variables of a function.

Range- the set of values that a given function can take as its argument varies.

8 0

3 years ago

The price of an item has been reduced by 85% . The original price was $45. What is the price of the item now.

Kobotan [32]

Answer:the price for the item now it's 6.75

Step-by-step explanation:

Origin price is 45=100% after reduction comes to 85% =x

X stands for the unknown value of the reduction therefore

45=100%

X=85%

45*.85=100x

After calculation you realised the value of x reduction value it's 38.25

Hence, original price is 45 your less from the value reduced 38.25 therefore the value of the item is 6.75

3 0

3 years ago

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl

Naya [18.7K]

Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
$\int_a^b f(r) |r'| dt$

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector $\sqrt{(x')^2 + (y')^2 + (z')^2}$

$\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt$

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3 dt

$\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du$

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
$=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\ \\ =\frac{1}{6} (14^{3/2} - 1)$

7 0

3 years ago