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3 years ago

What is the value of the function at x=−3?

Mathematics

2 answers:

kap26 [50]3 years ago

5 0

y=4 is the value y when x = - 3

lyudmila [28]3 years ago

3 0

It looks like y = 4. Why is the graph black?
You just move to that x value on the x axis then go to your graph (line) and what y - value is it across from?

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Evaluate the following expression. [4 + (12 - 2) ÷ 5] ÷ 7

Anika [276]

Answer:

[4 + (12 - 2) ÷ 5] ÷ 7

[4 + 6 / 5] / 7

[10 / 5] / 7

2 / 7

= 2/7 or 0.28571428571

7 0

2 years ago

Read 2 more answers

What is the value of x? Please help! Thanks!

leonid [27]

Answer:

x=75

Step-by-step explanation:

We can use proportions to solve this problem. Put the side of the small triangle over the side of the large triangle.

42 50

------ = ----------

42+63 50+x

42 50

------ = ----------

105 50+x

Using cross products

42(50+x) = 50*(105)

2100+42x = 5250

Subtract 2100 from each side

2100-2100+42x = 5250-2100

42x =3150

Divide by 42

42x/42 = 3150/42

x = 75

8 0

3 years ago

Use a paragraph, flow chart, or two-column proof to prove the angle congruency. Given: ∠ CXY ≅ ∠ BXY

Llana [10]

Answer:

See explanation

Step-by-step explanation:

Consider triangles CAX and BAX. In these triangles,

$\overline{AC}\cong \overline{AB}$ - given
$\angle CAX\cong \angle BAX$ - given
$\overline{AX}\cong \overline{AX}$ - reflexive property

By SAS postulate, $\triangle CAX\cong \triangle BAX$

Congruent triangles have conruent corresponding parts. So,

$\overline{CX}\cong \overline{BX}$

Consider triangles CXY and BXY. In these triangles,

$\overline{CX}\cong \overline{BX}$ - proven
$\angle CXY\cong \angle BXY$ - given
$\overline{XY}\cong \overline{XY}$ - reflexive property

By SAS postulate, $\triangle CXY\cong \triangle BXY$

Congruent triangles have conruent corresponding parts. So,

$\angle XCY\cong \angle XBY$

5 0

3 years ago

What is the area of triangle ABC? square units

andriy [413]

I think its 7. i know its late but im taking it now. myself

8 0

3 years ago

Read 2 more answers

Intergal of 8sin^4(x)dx

Romashka-Z-Leto [24]

$\int\limits8sin^4(x)dx =$

Take the constant out $\int\limits a*f(x)dx=a* \int\limitsf(x)dx$

$= 8 \int\limits sin^4(x)dx$

$\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx$

$= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)
$

Use the following identify : $sin^2(x) = \frac{1-cos(2x)}{2}$

$= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)$

$\int\limits 1dx = x$

$\int\limits cos(2x)dx = \frac{1}{2} sin(2x)$

Apply the sum rule :

$=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))$

Simplify :

$8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))$

Therefore add a constant to the solution:

$= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C$

hope this helps!

3 0

3 years ago