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3 years ago

If CDE ~ GDF, find ED

Mathematics

1 answer:

qaws [65]3 years ago

7 0

Answer:

Step-by-step explanation:

$\triangle CDE \sim \triangle GDF.. (given) \\\\\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\\therefore \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore \: 15 \times 4 = (x + 3)(3x + 1) \\ \\ \therefore \: 60 = 3 {x}^{2} + x + 9x + 3 \\ \\ \therefore 3 {x}^{2} + 10x + 3 - 60 = 0 \\ \therefore 3 {x}^{2} + 10x - 57 = 0 \\ \therefore 3 {x}^{2} + 19x - 9x - 57 = 0 \\ \therefore \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore \: (3x + 19)(x - 3) = 0 \\ \therefore \: 3x + 19 = 0 \: \: or \: \: x - 3 = 0 \\ \therefore \: x = - \frac{19}{3} \: \: or \: \: x = 3 \\ \because \: x \: can \: not \: be \: - ve \\ \therefore \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1 \\ \huge \red{ \boxed{ ED= 10}}$

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If a figure is dilated by a scale factor less than one, does the shape get larger or smaller?

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Solve x2 - 8x + 14 = 2x - 7

BabaBlast [244]

Answer:

x=7, x=3

Step-by-step explanation:

$x^{2} -8x+14=2x-7$

$x^{2} -8x-2x+14+7=0$

$x^{2} -10x+21=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=$ $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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$x=\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:21}}{2\cdot \:1}$

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negative

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$x=3$

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3 years ago

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Radda [10]

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Step-by-step explanation:

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Determine whether the set of vectors is a basis for ℛ3. Given the set of vectors , decide which of the following statements is t

schepotkina [342]

Answer:

(A) Set A is linearly independent and spans $R^3$ . Set is a basis for $R^3$ .

Step-by-Step Explanation

Definition (Linear Independence)

A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.

Definition (Span of a Set of Vectors)

The Span of a set of vectors is the set of all linear combinations of the vectors.

Definition (A Basis of a Subspace).

A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.

Given the set of vectors $A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)$ , we are to decide which of the given statements is true:

In Matrix $A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)$ , the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. $R^3$ has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans $R^3$ .

Therefore Set A is linearly independent and spans $R^3$ . Thus it is basis for $R^3$ .

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3 years ago

timama [110]

Answer:

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Step-by-step explanation:

The shape is a quarter circle with radius 10 units.

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area = 25π units^2

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