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3 years ago

Are you good at math or not good at math. Are you good at reading or are you not good at reading. Are tou good at both

Mathematics

2 answers:

son4ous [18]3 years ago

4 0

Answer:

im good at math to a certain point and im pretty good aT READING

Step-by-step explanation:

SpyIntel [72]3 years ago

3 0

Answer:

I am good at reading, but I am TERRIBLE at math..

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Look at the circle you created that has point C (the midpoint of AB) as its center and passes through point A. What can you say

Mice21 [21]

The inscribed angle on the circumference of the circle subtended by the diameter at the center is a right angle

What can be said about segment AB is; Segment AB is the diameter of circle with midpoint at C

What can be said about angle BDA is; Angle BDA is an inscribed angle of the circle with midpoint at C, subtended by the diameter AB at the center

The measure of angle BDA is 90°

What can be said about angle BEA is; Angle BEA is an inscribed angle of the circle with midpoint at C, subtended by the diameter AB at the center

The measure of angle BEA is 90°

The given diagram shows;

Circle with midpoint C, along AB, and radius AC

The diameter of circle C = AB

Circle with midpoint A, intersecting with circle C at points D and E

Tangents from point B on circle, C, intersect with circle A at points D and E

Required parameters;

What can be said about AB

The segment AB is a line that intersects the circle C at two points and also passes through the the point C which is the center of the circle C

Therefore, the segment AB is the diameter of the circle with center at C

What can be said about angle BDA

The angle BDA is the inscribed angle of circle C subtended by the points A and B which specifies the diameter of the circle with midpoint C

Therefore, the angle BDA is subtended by the diameter of the circle at the center

According to circle theorem, we have;

Angle subtended at the center = 2 × The angle subtended at the circumference

The angle subtended at the center by the diameter AB = 180° (Angle on a straight line)

Therefore;

The angle subtended at the center = 180° = 2 × Angle BDA

Angle BDA = 180°/2 = 90°

Angle BDA = 90°

What can be said about angle BEA

Similarly, we have;

The angle subtended at the center = 180° = 2 × Angle BEA

Angle BEA = 180°/2 = 90°

Angle BEA = 90°

Find out more about inscribed angles of a circle here:

brainly.com/question/17021375

7 0

3 years ago

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

3 years ago

Lori wants to distribute 35 peaches equally into baskets. she will use more then 1 but fewer than 10 baskets. how many baskets d

Ilya [14]

She needs either 5 baskets or 7 baskets.

3 0

3 years ago

Read 2 more answers

Free points help me 200÷2+200×5000000

bogdanovich [222]

Answer:

1000000100 hope it helps

5 0

3 years ago

Read 2 more answers

8) BRAINLIEST AND 15 POINTS! PLS HELP ASAP

algol13

Answer:

a.) Between 0.5 and 3 seconds.

Step-by-step explanation:

So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.

If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.

Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.

So, the answer is a.) between 0.5 and 3 seconds.

4 0

3 years ago