All squares are polygons.

Mathematics

2 answers:

grigory [225]3 years ago

6 0

True true false false false

ANEK [815]3 years ago

4 0

True
True
False
False
False

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Using algebra, find the point at which the line k(x) = 5x - 1 intersects with the line h(x) = -3x - 1 ?

vichka [17]

The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Given: k(x) = 5x - 1, h(x) = -3x - 1

We need to find the point(if any) at which these two lines k and h meets.

To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.

Therefore, k(x) = h(x)

=> 5x - 1 = -3x - 1

=> 8x = 0

=> x = 0

k(x=0) = 5 * 0 - 1 = -1

Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Know more about "point of intersection" problems here: brainly.com/question/16929168

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2 years ago

Convert 5 feet 4 inches to centimeters. (2.54 cm = 1 inch) *<br> PLEASE HELPPPP

hjlf

Answer:

162.56 cm

Step-by-step explanation:

also you can use a conversion calculator for faster results too

can i get brainliest plz

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2 years ago

If Ben invests $4500 at 4% interest per year, how much additional money must he invest at 5 1/2% annual interest to ensure that

pogonyaev

$Step \; 1: \; Assing \; variable \; for \; the \; unknown \; that \; we \; need \; to \; find.\\Let \; x \; be \; additional \; money \; invested$

$Step \; 2: \; Use \; the \; appropriate \; formula \; of \; simple \; interest \; set \; up \; an \; equation\\Interest \; earned \; each \; year = \; Amount \; invested \times annual \; rate \; of \; interest\\\\Amount \; invested \; in \; 4 \% \; rate \; of \; interest=\$4500\\Amount \; invested \; in \; 5\frac{1}{2}\% \; rate \; of \; interest= \$x\\Amount \; invested \; totally \; in \; 4\frac{1}{2}\% \; rate \; of \; interest = \$(4500+x)$

$Interest \; earned \; in \; 4\% \; interest \; rate\\ = \; 4500(4\%)=180\\Interest \; earned \; in \; 5\frac{1}{2}\% \; interest \; rate\\ = x(5\frac{1}{2}\%)=0.055x\\Interest \; earned \; totally \; in \; 4\frac{1}{2}\% \; interest \; rate = (4500+x)(4\frac{1}{2}\%)\\=0.045(4500+x)$

$\\\\So, \; the \; equation \; would \; be:\\180+0.055x=0.045(4500+x)$

$Step \; 1: \; Distribute \; 0.045 \; in \; the \; right \; side\\180+0.055x=202.5+0.045x\\\\Step \; 2: \; Subtract \; 0.045x \; and \; 180 \; on \; both \; sides \; to \; get \; x \; alone\\0.055x-0.045x=202.5-180\\\\Step \; 3: \; Combine \; Like \; Terms\\0.01x=22.5\\\\Step \; 4: Divide \; 0.01 \; on \; both \; sides\\\frac{0.01x}{0.01}=\frac{22.5}{0.01}\\\\Step \; 5: \; Simplifying \; fraction \; on \; both \; sides\\x=2250$