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3 years ago

Can someone please help me

Mathematics

1 answer:

mina [271]3 years ago

4 0

Answer:

The value of x is 7

AC congruent to DF because AC = 13.5 in. and DF = 13.5 in, EF congruent to BC because EF = 12 in. and BC = 12 in., ∠C congruent to ∠F. The triangles ACB and DEF congruent by the SAS Triangle Congruence Theorem only when x = 7

Step-by-step explanation:

<em>If two sides in a triangle congruent to two sides in other triangle and the including angles between the two sides are congruent in the triangles, then the two triangles are congruent by the SAS postulate of congruency</em>

In the given figure

→ If ∠C ≅ ∠F

∵ m∠C = m∠F

∵ m ∠ C = (2x + 14)°

∵ m∠F = (4x)°

→ Equate them to find x

∴ 4x = 2x + 14

→ Subtract 2x from both sides

∴ 2x = 14

→ Divide both sides by 2

∴ x = 7

→ In Δs ABC and DEF

∵ AC congruent to DF because AC = 13.5 in. and DF = 13.5 in

∵ EF congruent to CB because EF = 12 in. and CB = 12 in.

∵ ∠C congruent to ∠F

∴ The triangles ACB and DEF congruent by the SAS Triangle Congruence

Theorem only when x = 7

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Will upvote!!!

ipn [44]

B because 2^3=8
8^1/3=2

5 0

3 years ago

HH represents the height of the tree (in centimeters), t tt years since Renata moved in. H = 210 + 33 t H=210+33tH, equals, 210,

Jet001 [13]

Answer:

Therefore the rate of growth of the tree is 33 cm/ yr.

Step-by-step explanation:

Given that,

The height of the tree is

H= 210 +33t

Here H represent the height of tree in cm and t is time in year.

The rate of growth is the first order derivative of the H with respect to time.

i.e The rate of growth $=\frac{dH}{dt}$

∴H= 210 +33t

Differentiating with respect to t

$\frac{dH}{dt} =0 +33$

$\Rightarrow \frac{dH}{dt}=33$

Therefore the rate of growth of the tree is 33 cm/ yr.

7 0

3 years ago

NEED ANSWER QUICK PLSSSS

Gelneren [198K]

Answer:

CCC

Step-by-step explanation:

7 0

3 years ago

You are saving money for a summer camp that costs $1800. You have saved $500 so far, and you have 14 more weeks to save the tota

Arlecino [84]

Answer:

$92.86

Step-by-step explanation:

Since it's AT LEAST 1800, the inequality sign would be ≥ 1800

Equation:

500 + 14x ≥ 1800

14x ≥ 1800 - 500

14x ≥ 1300

x ≥ 92.86

6 0

3 years ago

Section 5.2 Problem 21:

Fittoniya [83]

Answer:

$y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$ (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation $am^2+bm+c=0$ where the values of $m$ are the roots:

$y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i$

Since the values of $m$ are complex conjugate roots, then the general solution is $y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]$ where $m=\alpha\pm\beta i$ .

Thus, the general solution for our given differential equation is $y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]$ .

To account for both initial conditions, take the derivative of $y(x)$ , thus, $y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]$

Now, we can create our system of equations given our initial conditions:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3$

$y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2$

We then solve the system of equations, which becomes easy since we already know that $C_1=3$ :

$-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}$

Thus, our final solution is:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$

7 0

2 years ago