Answer:
Solution given:
A triangle PQR is right angled at R, with hypotenuse{h}PQ=80cm
and
base[b]PR=60cm.
perpendicular [P]= QR
<u>by</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>
<u>h²</u><u>=</u><u>p²</u><u>+</u><u>b²</u>
80²=QR²+60²
QR²=80²-60²
QR=
QR=20
=52.9=53cm
<u>QR</u><u>=</u><u>5</u><u>3</u><u>c</u><u>m</u><u>.</u>
Answer:
14 MUST CHECK TO SEE IF IM CORRECT
Step-by-step explanation:
a+b>c
the two smallest numbers added together must be greater than the greatest number
ex: the triangle has three sides; 2,3,4 it could form a triangle because 2+3>4
SO HERE:
THE NUMBER MUST BE GREATER THAN 13 ADN THE SMALLEST WHOLE NUMBER GREATER THAN 13 IS 14
Answer:
XY = 18
Step-by-step explanation:
The figures are similar so the ratios of corresponding sides are equal, that is
=
, substitute values
=
( cross- multiply )
10 XY = 180 ( divide both sides by 10 )
XY = 18
The x-intercept is when y = 0
y = x - 5 → 0 = x - 5 → x = 5
The y-intercept is when x = 0
y = x - 5 → y = 0 - 5 → y = - 5
Answer: x-intercept is (5, 0), y-intercept is (0, -5)
Answer:
Numbers are 28 and -15
Step-by-step explanation:
Mark first number X, and second number Y, then:
x-y=43
and
x+y=13
We get system of two equations. Solve it by add both them together:
x-y+x+y=43+13
2x=56
x=28 we get first number
x+y=13
28+y=13
y=13-28
y=-15