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3 years ago

12

Solve this equation by Euclid's application of areas: x^2 + 12x = 64.

1 answer:

Karolina [17]3 years ago

7 0

Answer:

4

Step-by-step explanation:

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Felicia wants to build a kite with the shape shown. If AC is 60 cm, how many centimeters are in the length of BD?

Kay [80]

Answer:

Step-by-step explanation:

By applying tangent rule in the given right triangle AOB,

tan(30°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

$\frac{1}{\sqrt{3}}=\frac{BO}{OA}$

$OA=BO(\sqrt{3})$

By applying tangent rule in the given right triangle BOC,

tan(60°) = $\frac{OC}{BO}$

OC = BO(√3)

OA + OC = AC

$BO(\sqrt{3})+BO(\sqrt{3}) =60$

2√3(BO) = 60

BO = 10√3

OC = BO(√3)

OC = (10√3)(√3)

OC = 30

By applying tangent rule in right triangle DOC,

tan(60°) = $\frac{OD}{OC}$

OD = OC(√3)

OD = 30√3

Since, BD = BO + OD

BD = 10√3 + 30√3

BD = 40√3

≈ 69.3

6 0

3 years ago

What is the solution to this system of linear equations?

nataly862011 [7]

Step-by-step explanation:

the question answer a should be

3 0

2 years ago

What is the length of ab and bc ? a= (1,-3) b= (3,-1) C= (5,-3)

ddd [48]

Answer:

the length is (1,3

Step-by-step explanation:

the length is 3,1

5 0

3 years ago

I need to solve x pls

Marat540 [252]

Answer:

I think x is 38 but it depend on how to solve it

8 0

2 years ago

solve the given matrix equation for X. Simplify your answers as much as possible. (In the words of Albert Einstein, "Everything

vova2212 [387]

a.

$XA^{-1}=A^3$

$(XA^{-1})A=A^3A$

$X(A^{-1}A)=A^4$

$X=A^4$

b.

$AXB=(BA)^2$

$A^{-1}(AXB)B^{-1}=A^{-1}(BA)^2B^{-1}$

$(A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}$

$X=A^{-1}(BA)^2B^{-1}$

c.

$(A^{-1}X)^{-1}=(AB^{-1})^{-1}(AB^2)$

$X^{-1}A=(BA^{-1})(AB^2)$

$X^{-1}A=B(A^{-1}A)B^2$

$X^{-1}A=B^3$

$(XX^{-1})A=XB^3$

$XB^3=A$

$X(B^3(B^3)^{-1})=A(B^3)^{-1}$

$X=A(B^3)^{-1}$

d. Not totally sure what the equation is supposed to be, but I guess it's

$ABXA^{-1}B^{-1}=A$

$ABX(BA)^{-1}=A$

$((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)$

$X=(AB)^{-1}A(BA)$

$X=(B^{-1}A^{-1})A(BA)$

$X=B^{-1}(A^{-1}A)(BA)$

$X=(B^{-1}B)A$

$X=A$

7 0

4 years ago