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irga5000 [103]
3 years ago
11

Hey can you please help me posted picture of question

Mathematics
2 answers:
Mumz [18]3 years ago
5 0
1. X-3=0
Add 3 to both sides to cancel the negative 3.
Your answer is a positive 3
2. X-4=0
Add 4 to both sides to cancel the negative 4
Your answer is 4 option C is correct
Andre45 [30]3 years ago
4 0
X = 3 and x = 4 will make the polynomial = 0

answer
C. 3 and 4
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6. The distance from Tory's home to the mall is 14 km. The distance from her home
Minchanka [31]

Answer: 6 km to 22 km !

Step-by-step explanation:

the distances from the school mall to bus station could be, in the worst case, 14+8=22 km apart, and in the best case 14-8=6 km apart

so the range is 6 km to 22 km !

8 0
2 years ago
H(x) = 5х+2.0х 3 - 45х-180х.<br> factor the polynomial
poizon [28]
The correct answer is 2x^3-40x-180
4 0
3 years ago
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

8 0
3 years ago
Sketch the graph of each line y=-9x-5
ycow [4]
I can’t answer it without a picture of the graph
5 0
3 years ago
Share 160 sweets in the ratio 8:12
katrin [286]

let's divide the whole by the sum of its ratios, namely let's 160 ÷ (8+12), and then distribute accordingly.

\bf \textit{160 at 8:12 ratio}\qquad \qquad \cfrac{8\cdot \frac{160}{8+12}}{12\cdot \frac{160}{8+12}}\implies \cfrac{8(8)}{12(8)}\implies \cfrac{64}{96}

5 0
3 years ago
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