What is an equation in a slop-intercept from for the line that passes through the points (1,-3) and (3,1) ?
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The shop sold 36 sundaes and 28 banana spilts
<em><u>Solution:</u></em>
Let "a" be the number of sundaes sold
Let "b" be the number of banana spilts sold
Cost of 1 sundaes = $ 2
Cost of 1 banana spilt = $ 3
<em><u>On a hot summer day, the shop sold 8 more sundaes than banana splits and made $156</u></em>
Therefore,
Number of sundaes sold = 8 + number of banana spilts sold
a = 8 + b ------- eqn 1
<em><u>Also, given that they made $ 156</u></em>
<em><u>Therefore, we frame a equation as</u></em>:
Number of sundaes sold x Cost of 1 sundaes + number of banana spilts sold x Cost of 1 banana spilt = 156
2a + 3b = 156 ---------- eqn 2
<em><u>Substitute eqn 1 into eqn 2</u></em>
2(8 + b) + 3b = 156
16 + 2b + 3b = 156
16 + 5b = 156
5b = 156 - 16
5b = 140
<h3>b = 28</h3><em><u>Substitute b = 28 in eqn 1</u></em>
a = 8 + 28
<h3>a = 36</h3>Thus the shop sold 36 sundaes and 28 banana spilts
See the picture attached to better understand the problem
we know that
If two secant segments are drawn to a <span>circle </span><span>from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
</span>so
jl*jk=jn*jm------> jn=jl*jk/jm
we have
<span>jk=8,lk=4 and jm=6
</span>jl=8+4----> 12
jn=jl*jk/jm-----> jn=12*8/6----> jn=16
the answer is
jn=16
we know that
If two secant segments are drawn to a <span>circle </span><span>from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
</span>so
jl*jk=jn*jm------> jn=jl*jk/jm
we have
<span>jk=8,lk=4 and jm=6
</span>jl=8+4----> 12
jn=jl*jk/jm-----> jn=12*8/6----> jn=16
the answer is
jn=16