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alexgriva [62]
2 years ago
10

Could someone help me with this problem!:)

Mathematics
1 answer:
kow [346]2 years ago
4 0

Answer:

i think it's 9 just correct me if I'm wrong

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Solve 10x3^-40x2^+40x=0
suter [353]

Solve for x by simplifying both sides of the equation, then isolating the variable.

x=0

plz mark me as brainliest :)

3 0
3 years ago
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In Mayfield, the taxi fare is $3.50 for the first mile and an additional S.50 for each mile after the first.You plan to give the
defon

We were told that the taxi fare is $3.50 for the first mile and an additional S.50 for each mile after the first. Assuming that you ride x additional miles after the first, the total cost would be

3.5 + 0.5x

Given that you plan to give the driver a $3 tip for your 13-mile taxi ride, the equation representing the cost of the ride, c would be

Remember the cost of the first mile is $3.5. Since the total number of miles is 13, the number of dditional miles would be 13 - 1 = 12 miles. The equation would be

c = 3.5*1 + 0.5 * 12 + 3

= 3.5 + 6 + 3

= $12.5

The cost of the ride would be $12.5

5 0
1 year ago
PLEASEEEEEEE answer I've been trying forever. :(
SCORPION-xisa [38]
Which question do you need help with
7 0
2 years ago
Read 2 more answers
What is the answer to this problem
Yuliya22 [10]
From the given values of x and y, the relation is linear and can be represented algebraically as the formula y=-6x+4. At x=8, y=-6*8+4=-48+4=-44
8 0
2 years ago
Graph the polygon and it’s image after a dilation centered C with scale factor k
givi [52]
Explanation<h2>Graphing the polygon</h2>

We plot and join the given ordered pairs.

<h2>Graphing the image of the polygon after the dilation</h2>

The formula of dilation when it is not centred at the origin is:

\begin{gathered} (x,y)\rightarrow(k(x-a)+a,k(y-b)+b) \\ \text{ Where} \\ k\text{ is the scale factor} \\ (a,b)\text{ is the center of the dilation} \end{gathered}

Then, we can find the coordinates of the image:

\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(7-(-2))-2,\frac{2}{3}(1-4)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(7+2)-2,\frac{2}{3}(-3)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(9)-2,\frac{2}{3}(-3)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(6-2,-2+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(4,2) \end{gathered}\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(4-(-2))-2,\frac{2}{3}(4-4)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(4+2)-2,\frac{2}{3}(4-4)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(6)-2,\frac{2}{3}(0)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(4-2,0+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(2,4) \end{gathered}

\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1-(-2))-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1+2)-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(3)-2,\frac{2}{3}(9)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(2-2,6+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(0,10) \end{gathered}

4 0
1 year ago
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