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Rashid [163]
3 years ago
6

Which of the following sets show all the numbers from the set {3, 4, 5, 6} that are part of the solution to the equation 5x + 4

< 29?
{4, 5, 6}
{3, 4 , 5}
{3, 4 }
{3}​
Mathematics
2 answers:
BartSMP [9]3 years ago
8 0

Answer:

3,4

Step-by-step explanation:

Vesna [10]3 years ago
4 0

Replace x with the possible answers and solve to see which ones make the equation true:

5(3) = 15 + 4 = 19. 19 is < 29 so 3 works.

5(4) = 20 + 4 = 24. 24 is < 29, so 4 also works.

5(5) = 25 + 4 = 29. 29 is not less than 29 so doesn't work.

Only 3 and 4 will work.

The answer would be {3,4}

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Among 1,200 Chinese tourists who are visiting Nepal,45 of them already visited India,40 have visited Pakistan and 15 have visite
SOVA2 [1]

30 visited nepal only 25 visited Pakistan only. 15 tourists at the intersection of the two circles and 1130 have not visited either

8 0
3 years ago
Read 2 more answers
How many people swam fewer than 30 laps?
Bas_tet [7]

Answer:

3 people

Step-by-step explanation:

Reading a stem & leaf plot can be hard!

Remember in the middle or whichever side is single digits divided by the line is your tens place values.

The 2 has two values on the other side, giving us 23 & 29 laps.

The 1 has one value on the other side, giving us 11 laps.

The 0 has no other value, so there is no data for that row.

There were 3 data points below 30, & these were 11, 23, & 29.

Hope this helped. :)

5 0
2 years ago
Braden biked 10 8/9 miles on Monday. On Tuesday, he biked 3/ 4 as far as Monday. How many total miles did he bike over the two d
torisob [31]

Answer:

19 1/18 miles.

Step-by-step explanation:

For Tuesday's ride we need to multiply 10 8/9 by 3/4.

10 8/9 = (9*10+8) / 9 =  98/9 miles.

98 / 9  * 3/4

= 98/3 * 1/4

49/6

= 8 1/6 miles.

Total miles over 2 days = 10 8/9 + 8 1/6

= 18 + 8/9 + 1/6

= 18 +  48/54 + 9/54

= 18 57/54

= 19 3/54

= 19 1/18 miles.

5 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

4 0
3 years ago
Write the quadratic equation in standard form and then choose the value of "b." (2x - 1)(x + 5) = 0
Sunny_sXe [5.5K]
If we were to foil
after experieence

we know
ax²+bx+c=0

and
in form
(ax+b)(cx+d)=0
if we expand it, we get
acx²+bcx+adx+bd=0
or
(ac)x²+(bc+ad)x+(bd)=0
compare to
ax²+bx+c=0
we notice that the middle terms (x terms) are

b=(bc+ad)
so

in form
(2x-1)(1x+5)
b=bc+ad=(-1*1+2*5)=-1+10=9

b=9
or you could just expand it
3 0
3 years ago
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