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3 years ago

Graph f(x) = 4[x] + 5

Mathematics

1 answer:

Elena L [17]3 years ago

6 0

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Justice hopes to build a fence around his garden he measures and determined that he needs 24 feet of fencing when he arrives at

Sunny_sXe [5.5K]

Answer:

It will cost $72 Justice to fence in his garden.

Step-by-step explanation:

Total length of the fencing around the garden = 24 feet

Cost of fencing per inch = $0.25

Now, as we know 1 foot = 12 inches

⇒ 24 feet = 24 x 12 inch = 288 inches

⇒ 24 feet = 288 inches

Hence, the total length of fencing = 288 inches.

Now, Total cost of Fencing

=Total length of fencing x Cost of fencing par inch

= 288 inches x ($0.25)

= $ 72

or, the total cost of fencing = $72

Hence, it will cost $72 Justice to fence in his garden.

7 0

2 years ago

A submarine travels 6.6 km due West from its base and then turns and travels due South for 3.9

geniusboy [140]

Answer:

$7.7$ km

Step-by-step explanation:

The submarine's path from its base forms a right triangle when its final position is "connected" to the base. We know that the right triangle has legs of $6.6$ km and $3.9$ km, and we need to find the length of its hypotenuse. To do so, we can use the Pythagorean Theorem, which states that in a right triangle, $a^{2}+b^{2} =c^{2}$ , where $a$ and $b$ are the lengths of the triangle's legs and $c$ is the length of the triangle's hypotenuse. In this case, we know what $a$ and $b$ are, and we need to solve for c, so after substituting the given values of $a$ and $b$ into $a^{2}+b^{2} =c^{2}$ to solve for c, we get:

$a^{2}+b^{2} =c^{2}$

$6.6^{2} +3.9^{2} =c^{2}$ (Substitute $a=6.6$ and $b=3.9$ into the equation)

$43.56+15.21=c^{2}$ (Evaluate the squares on the LHS)

$58.77=c^{2}$ (Simplify the LHS)

$c^{2}=58.77$ (Symmetric Property of Equality)

$\sqrt{c^{2} } =\sqrt{58.77}$ (Take the square root of both sides of the equation)

$c=7.7,c=-7.7$ (Simplify)

$c=-7.7$ is an extraneous solution because you can't have negative distance, if that makes sense, so therefore, the submarine is approximately $7.7$ km away from its base. Hope this helps!