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3 years ago

AC=CE and D is the midpoint of CE. If CE = 10x+18 DE = 7x-1 and BC = 9x-3 find AB

Mathematics

1 answer:

Mazyrski [523]3 years ago

8 0

We know the CE = 10x + 18 and DE = 7x - 1 and D is the midpoint of CE

⇒ CE = 2 * DE

⇒ 10x + 18 = 2 * (7x - 1)

⇒ 10x + 18 = 14x - 2

⇒ -4x = -20

⇒ x = 5

Now we also know that BC = 9x - 3 and AC = CE = 10x + 18

⇒ AB = AC - BC

⇒ AB = 10x + 18 - (9x - 3)

⇒ AB = x + 21

Substituting the value of x,

⇒ AB = 5 + 21

⇒ AB = 26

Hence, the value of AB is 26 units.

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State whether the equation is true, false or open? 4y+8=6y+3

g100num [7]

Answer:

Open

Explanation:

An open equation is an equation that has variables and CAN be solved

Example: 2x = 4 ......> can be solved giving ......> x = 2

A false equation is one where both sides can NEVER be equal

Example: 15 = 2(3) + 1 ..........> 15 can never be equal to 7

A true equation is one having no variables and both sides are ALWAYS equal

Example: 2(3) + 1 = 2(2) + 3 ........> 7 will always be equal to 7

Now, the given equation is:

4y + 8 = 6y + 3

Let's try to solve is:

4y + 8 = 6y + 3

6y - 4y = 8 - 3

2y = 5

y = 2.5

Therefore, the given equation contains a variable and can be solved which means that it is an open equation

Hope this helps :)

8 0

3 years ago

Read 2 more answers

BRAINLIEST FOR RIGHT ANSWER!!

julsineya [31]

Answer: ${\boxed{h(x+4)=\frac{9+3x}{8+x}}}$

Concept:

In a function f(x), it represents a function in terms of x or for x. Therefore, to find the values in the function, substitute values within the parenthesis to solve.

Solve:

Given information

$h(x)=\frac{-3+3x}{4+x}$

Need to find

$h(x+4)$

Substitute (x + 4) to the position of x

$h(x+4)=\frac{-3+3(x+4)}{4+(x+4)}$

Distributive property on the numerator

$h(x+4)=\frac{-3+3x+12}{4+(x+4)}$

Combine like terms

$h(x+4)=\frac{-3+12+3x}{4+4+x}$

$\boxed{h(x+4)=\frac{9+3x}{8+x}}$

Hope this helps!! :)

Please let me know if you have any questions

5 0

2 years ago

Please help :( really stuck on this and don’t understand please attempt to do it .

kenny6666 [7]

Answer) That graph is not a function
Explanation) The graph that you provided is not a function. It does not pass the vertical line test. The vertical line test is when you draw a vertical line (l) at any point on the graph and it should touch 1 or less parts of the graph. If you put the line at x=1, the vertical line only touches the graph at (1,8.5) but if you put the line at x=5, it touches (5,1) and (5,8.5) so it does not pass the test. You should be able to put the line anywhere and have it touch ONLY 1 point. There cannot be multiple of the same x values.

3 0

2 years ago

3/4 W + 8 = 1/3 w – 7

Montano1993 [528]

Answer:

Step-by-step explanation:

$\dfrac{3}{4}W+8=\dfrac{1}{3}W-7\qquad\text{subtract 8 from both sides}\\\\\dfrac{3}{4}W+8-8=\dfrac{1}{3}W-7-8\\\\\dfrac{3}{4}W=\dfrac{1}{3}W-15\qquad\text{multiply both sides by}\ LCM(4,\ 3)=12\\\\12\!\!\!\!\!\diagup^3\cdot\dfrac{3}{4\!\!\!\!\diagup_1}W=12\!\!\!\!\!\diagup^4\cdot\dfrac{1}{3\!\!\!\!\diagup_1}W-(12)(15)\\\\(3)(3W)=(4)(1W)-180\\\\9W=4W-180\qquad\text{subtract}\ 4W\ \text{from both sides}\\\\9W-4W=4W-4W-180\\\\5W=-180\qquad\text{divide both sides by 5}\\\\\dfrac{5W}{5}=\dfrac{-180}{5}\\\\W=-36$

7 0

3 years ago

42:28

gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) $\overline{ED}\cong \overline{FD}$

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

$\\ \overline{EF}\cong \overline{FD}$ (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD = 60°

ΔEFD is an equilateral triangle as all interior angles are equal

$\\ \overline{EF}\cong \overline{FD}\cong \overline{ED}$ (definition of equilateral triangle)

$\overline{ED}\cong \overline{FD}$

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

$\overline{CF}\cong \overline{CD}$ (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0

3 years ago