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3 years ago

There is a part c Part c What is the length and width of each section?

Mathematics

1 answer:

mr Goodwill [35]3 years ago

3 0

Answer:

Part A: Width is 1/4 yard

Part C: Each section: L = 1/9 yard, W = 1/8 yard

Step-by-step explanation:

Part A:

The area is given as 1/12 yard² and the length is given as 1/3 yard. The area of a rectangle is A = LW, plug in the given values and solve...

1/12 = (1/3)W

3/12 = W (multiply both sides by 3 to get rid of the fraction on the right)

1/4 yard = W (reduce the fraction on the left)

Part C: The width is 1/4, and there are 2 squares along the width, so each section is

1/4 ÷ 2 = (1/4)*(1/2) = 1/8 yards wide

The length is 1/3 and there are 3 sections along the length, so each section is

1/3 ÷ 3 = (1/3)*(1/3) = 1/9 yards long

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You and your friend live 12 miles apart. you leave home at the same time and travel ttoward eacth other.you walk at the rate of

mr Goodwill [35]

Step 1: Assign variable for the unknown that you need to find.

$d_w \; \rightarrow \; Distance \; you \; traveled \; by \; walk \; in \; miles \\ d_c \; \rightarrow \; Distance \; traveled \; by \; your \; friend \; using \; bicycle \; in \; miles \\ t \; \rightarrow \; Time \; Taken \; in \; hours$

Step 2: List the given details

Distance between you and your friend is 12 miles. It means the total distance traveled by you by walk and your friend using bicycle when both travel towards each other to meet in 't' hours

Equation 1:

$d_w+d_c=12$

Your Walking rate and your friends Cycling rate is given below:

$r_w=4mph\\r_c=11mph$

Step 3: Set up an equation using the relationship between distance, time and rate.

$Distance=Rate \times Time$

$d_w=r_w \times t=4t \\\\ d_c = r_c \times t = 11t$

Substituting the above in equation 1

$4t+11t=12 \\ Combining \; Like \; Terms \\\\ 15t=12 \\ Dividing \; 15 \; on \; both \; sides \\\\ 15t \div 15=12 \div 15 \\\\ t=0.8 hours$

Distance traveled by you by walking from home and distance travelled by your friend in cycle in 0.8 hours is given below:

$d_w=4t=4(0.8)=3.2miles \\\\ d_c=11t=11(0.8)=8.8miles$

8 0

3 years ago

Find the critical points, domain endpoints, and local extreme values for the function

aliina [53]

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = \frac{10\cdot x \cdot (x+3)-5\cdot x^{2}}{25\cdot (x+3)^{2}}$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^{2} = 0$

$10\cdot x^{2}+30\cdot x -5\cdot x^{2} = 0$

$5\cdot x^{2} + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)

8 0

3 years ago

Simplify -12x + 5x.<br><br> -17 x<br> -7<br> -7 x

Blababa [14]

Answer:

-7x

Step-by-step explanation:

==> -12x + 5x.

==> -12+5xx

==> -7x

8 0

2 years ago

Read 2 more answers

The strength of a cable is proportional to the square of its diameter. If a 2-cm cable will support 800kg, how much will a 3-cm

xz_007 [3.2K]

Strength [is proportional to] d^2
strength1/(d1)^2 = strength2/(d2)^2
800kg/(2cm)^2 = strength2/(3cm)^2
strength2 = 800 kg * (3 cm)^2/(2 cm)^2
strength2 = 800 kg * 3^2/2^2
strength2 = 800 kg * 9/4
strength2 = 1800 kg

3 0

3 years ago

Mary has a recipe for cranberry bread. she uses 6 and 2/3 cups of flour to make 2 loaves of cranberry bread.

sweet [91]

Answer: it’s the third one

Step-by-step explanation:

It is

8 0

2 years ago