Answer:
11/30 or 0.36
Step-by-step explanation:
Answer:
In order to have ran 33 miles, Bobby would have to attend <em>32 track practices.</em>
Step-by-step explanation:
Solving this problem entails of uncovering the amount of track practices Bobby must attend in order to have ran 33 miles. Start by reading the problem carefully to break down the information provided.
You can see that Bobby has already ran one mile on his own. This is important to remember for later. The problem also states that he expects to run one mile at every track practice.
Setting up an equation will help us solve. Here is how we could set up the equation:
(<em>amount of miles already ran</em> = 1) + (<em>number of track practices</em> = x) = (<em>total miles to run</em> = 33)
1 + x = 33
The equation is now in place. You can solve this, or isolate <em>'x',</em> by using the subtraction property of equality. This means we will subtract one from both sides of the equation, thus isolating the variable.
1 + x = 33
1 - 1 + x = 33 - 1
x = 32
The variable is the only term left on the left side of the equation. This means Bobby must attend track practice <em>32 times</em> in order to have ran 33 miles.
So in this case, the solution is where the graphs intersect. Put the equations in y-intercept form.
-4x+y=-6
y=4x-6
8x-2y=14
-2y=14-8x
y=4x-7
These lines have the same slope, which is 4. Therefore, they are parallel. Parallel lines never intersect, so there is NO SOLUTION.
Answer:
a- 10x^2 - 4x + 12
b- no, the first equation is −1/4⋅(5x^2+6x−24) and the second is -1/4⋅(10x^2+7x-24)
Step-by-step explanation:
m6th w6y
BRAINLIEST??
Answer:
m<GFA = 110
Step-by-step explanation:
1. ABCD - parallelogram Definition of a parallelogram
(AB ll CD) (AD ll BC)
2. m<B + m<C = 180 Consectuive angles in a
110 + m<C = 180 parallelogram are supplementary
m<C = 70
3. m<GCB = 1/2 m<C Definition of angles bisector
m<GCB = 70
4. m<B = m<D = 110 Opposite angles in a
parrallelogram are congruent
5. m<CDG = 1/2 m<D Defintion of an angle bisector
m<CDG = 55
6. m<GCB+m<CDG+m<CGD=180 Sum of anlges in a triangle (ΔCDG)
70 + 55 + m<CGD = 180
125 + m<CGD = 180
m<CGD = 55
7. m<CGD + m<DGF = 180 Linear pair, supplmentary angles
55 + m<DGF = 180
m<DGF = 125
8. m<C = m<A = 70 Opposite angles in a paralellogram
are congruent
9. m<ADG = 1/2m<D Definiton of an angle bisector
m<ADG = 55
10.m<ADG+m<DFG+m<GFA+m<A=360 Sum of angles in quadrilateral
55 + 125 + m<GFA + 70 = 360 DGFA
m<GFA + 250 = 360
m<GFA = 110