All categories

3 years ago

Can I have help with this problem?

Mathematics

1 answer:

elena-s [515]3 years ago

7 0

Answer:

angles that measure 88° are ∡'s 1, 8, and 3

Step-by-step explanation:

m∡1 = 88° because it is a vertical angle with the 88° angle in blue

m∡ 8 = 88° because it is corresponding with the 88° angle in blue

m∡3 = 88 because it is an alternate-interior angle with the 88° angle in blue

You might be interested in

Hello, can someone please help me on solving this kind of question?

jek_recluse [69]

Answer: so i think for part a , the first term is a subscrpit 0 which is 10. Then the formula for any term in the sequence is Ao x R ^n. So R =2 so

1st term = 10 ( 2^0)= 1

2nd term = 10 (2^1) =20

3rd term = 10 (2^2)= 40

4th = 80

5th = 160

Part B

The limit as n approaches infinity for a sub n when R is 0, 1 and 2. THE LIMIT FOR R of 0, 1 and 2 is infinity.

Part C...i will have to research...sorry

Step-by-step explanation:

8 0

3 years ago

Find the part of the number.<br><br><br> * Find 3 1/16 of 32.

oee [108]

Answer:

Step-by-step explanation:

6 0

3 years ago

What the answer asap

Elan Coil [88]

20 qt = 80 c is ur answer. Hope this helps

3 0

3 years ago

Read 2 more answers

A semicircle with a diameter of 8 meters is shown. A smaller semicircle with a diameter of 4 meters is cut out of the larger sem

Lena [83]

Answer:

here

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

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