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elixir [45]
3 years ago
11

Teddy was having trouble fitting into his clothes due to weight gain. In January, he weighed 180 lbs 6 oz. He gained 3 lbs 9 oz,

5 lbs 12 oz, and 6 lbs 11oz in the next three months. How much does Teddy weigh now?

Mathematics
1 answer:
barxatty [35]3 years ago
7 0
He weighs 196 lbs and 6 ounces. if you add the pounds (180 to 3, 5, 6)you get 194. then you add he ounces, and remember that there are 16 ounces in a pound and 8 ounces in a cup, (6 to 9, 12, 11) you get 2lbs and 6 ounces. then you add the 2 extra pounds to the 194 and you get, all together, 196 lbs and 6 ounces.
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Factor 105^2-8^2 into two prime factors
Nonamiya [84]
105² - 8² = (105-8)(105+8) = 97 * 113
8 0
3 years ago
write an equation of a line that is perpendicular to the given line and that passes through the given point (4,-6); m=3/5
tresset_1 [31]

Answer:

y=-\frac{5}{3} x-\frac{42}{5}

Step-by-step explanation:

Given the slope and another point, simply plug them into the point-slope formula to find your y-intercept.

y-y1=m(x-x1)\\y-(-6)=\frac{3}{5} (x-4)\\y+6=\frac{3}{5} x-\frac{12}{5} \\y=\frac{3}{5} x-\frac{42}{5}

Now that we've found your y-intercept, we have the original equation. To find the perpendicular equation, you need the opposite reciprocal of your slope.

To find the 'opposite,' change your slope's sign. Since your slope is positive \frac{3}{5}, the opposite is -\frac{3}{5}.

To find the 'reciprocal,' flip your fraction. This will make your slope -\frac{5}{3}.

Your final equation is:

y=-\frac{5}{3} x-\frac{42}{5}

7 0
3 years ago
7 1/2 divided by 3/4
creativ13 [48]

Answer:

10

Step-by-step explanation:

8 0
3 years ago
Someone please help me with this
skelet666 [1.2K]

Answer:

see explanation

Step-by-step explanation:

(a)

f(0)

Since x = 0 < 5, then

f(0) = x + 4 = 0 + 4 = 4

(b)

f(6)

Since x = 6 meets the condition 5 ≤ x < 7, then

f(6) = 8

5 0
3 years ago
the line joining A(a, 3) to B(2 -3) is perpendicular to the line joining C(10,1) to B. The value of a is?
RideAnS [48]
Well, first off, let's find what is the slope of BC

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;B&({{ 2}}\quad ,&{{ -3}})\quad &#10;%   (c,d)&#10;C&({{ 10}}\quad ,&{{ 1}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2}&#10;\\\\\\&#10;\cfrac{4}{8}\implies \cfrac{1}{2}

now, a line perpendicular to that one, will have a negative reciprocal slope, thus

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\&#10;slope=\cfrac{1}{{{ 2}}}\qquad negative\implies  -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}

now, we know the slope "m" of AB is -2 then, thus

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;A&({{ a}}\quad ,&{{ 3}})\quad &#10;%   (c,d)&#10;B&({{ 2}}\quad ,&{{ -3}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2}&#10;\\\\\\&#10;\cfrac{-6}{2-a}=-2\implies -6=-4+2a\implies -2=2a\implies \cfrac{-2}{2}=a&#10;\\\\\\&#10;-1=a
6 0
3 years ago
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