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3 years ago

Does the point (5, 7) satisfy the inequality y < 7x + 10?

Mathematics

1 answer:

Anna71 [15]3 years ago

3 0

Answer:

yes

Step-by-step explanation:

y < 7x + 10

Substitute the point into the inequality and see if it is true

7 < 7*5 + 10

7 < 35+10

7 < 45

This is true so the point is a solution

You might be interested in

Factor completely.<br>28 - 7x^2

FrozenT [24]

Answer:

7(4 - x^2)

Step-by-step explanation:

Well you just factor out 7, since that is the GCF. 7(4 - x^2)

Hope this helps

8 0

3 years ago

Two miners are trying to push a stationary mine cart filled with gold. Miner A is exerting 25 N while miner B is exerting 20 N.

Oliga [24]

Answer:

b. 45 N

Step-by-step explanation:

The sum of the forces exerted by the two miners = 25 + 20

= 45 N

Since after applying a force of 45 N by the two miners the cart remained stationary, it implies that a resistance force of at least 45 N acts against their push. Thus, the force exerted by the road could be equal to 45 N or greater than 45 N.

Probably, if the sum of the forces applied is 46 N, the cart would move. Which depends on the value of resistance force applied by the road. Thus, the appropriate answer from the given option is B.

Since the force applied on the cart equals the resistance force exerted by the road, the cart would not move.

6 0

3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

Mel uses one fourth of a yard of ribbon to make a hair bow how many hair bows can mel make with 1 yard of ribbon

luda_lava [24]

Answer:

4 hair bow

Step-by-step explanation:

if 1/4 can make 1 hair bows you can make 4 hair bows

, it's simple 4/1=4

6 0

3 years ago

Read 2 more answers