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2 years ago

SSA triangles always have two solutions. (A.) True (B). False

Mathematics

2 answers:

scoray [572]2 years ago

8 0

Answer:

No, Ssa triangle always not have two solutions

Step-by-step explanation:

the solution can be one, two or there will be no dolution. So this statement is false.

Marina86 [1]2 years ago

5 0

The answer is false. hopefully i helped

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Find X : log2+ logx=1

Dominik [7]

$log2+logx=1;\ D:x\in\mathbb{R^+}\\\\log(2\ \cdot\ x)=log10\iff2x=10\ \ \ /:2\\\\x=5\in D\\\\Solution:x=5.$

8 0

3 years ago

Read 2 more answers

(HELP PLEASEE) The area of a rectangular swimming pool is 2x^2+13x-7. Find the dimensions of the swimming pool.

bearhunter [10]

Answer:

length =

$2x - 1$

width =

$x + 7$

Step-by-step explanation:

here we go,

=》

$2x {}^{2} + 13x - 7$

=》

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

=》

$2x(x + 7) - 1(x + 7)$

=》

$(2x - 1)(x + 7)$

here, we got

the dimensions as :-

$(2x - 1) \: \: and \: \: (x + 7)$

7 0

2 years ago

PLEASE HELPPPPP. URGEENNTT!!! I DONT KNOW HOW TO SOLVE THIS

Oksi-84 [34.3K]

Answer:

A = 58.7 degrees

B = 66.9 degrees

C = 34.1 degrees

Step-by-step explanation:

For <A:

Tan A = $\frac{opposite}{adjacent}$

Tan A = $\frac{23}{14}$

Tan A = 1.6

A = $Tan^{-1} 1.6$

A = 58.7 degrees

For <B:

Sin B = $\frac{opposite}{hypotenuse}$

Sin B = $\frac{23}{25}$

Sin B = 0.92

B = $Sin^{-1} 0.92$

B = 66.9 degrees

For <C:

Sin C = $\frac{opposite}{hypotenuse}$

Sin C = $\frac{14}{25}$

Sin C = 0.56

C = $Sin^{-1}0.56$

C = 34.1 degrees

6 0

3 years ago

PLEASE HELP ME (40 POINTS)

CaHeK987 [17]

For each equation give X a value and solve for Y, do this at least two times for each equation.

Plot those dots on a graph to draw two lines.

The solution is where the two lines cross.

You will see the lines cross at (-1,-2)

7 0

3 years ago

Three consecutive vertices of a parallelogram are (– 2, 1); (1, 0) and (4, 3). Find the coordinate of the fourth vertex and find

antiseptic1488 [7]

Answer:

D(1,4) - the fourth vertex, the area is equal to 12

Step-by-step explanation:

A(2,-1), B(1,0),C(4,3), They are consecutive, we need to find the point D, find the midpoint of AC (O)

xo= (xA+xc)/2 x0= (-2+4)/2=1

y0=(yA+yC)/2 y0= (1+3)/2=2

0(1,2)

O is the midpoint of BD, so let's find the vertex D

X0= (xB+xD)/2 1= (1+xD)/2 xD=1

y0= (yB+yD)/2 2= (0+yD)/2 yD=4

D(1,4)

the vector DA is (-2-1, 1-4)= (-3,-3)

the vector DC is (4-1, 3-4)= (3,-1)

The modul of DC is sqrt ((-3)^2+(-3)^2)= 3*sqrt2(the length of the side Dc)

The modul of DA is sqrt (3^2+(-1)^2)= sqrt10(the length of the side DA)

cosD= (-3*3+(-3)(-1))/3*sqrt2*sqrt10=-6/3*2*sqrt5=-1/sqrt5

sinD= sqrt (1- 1/5)=2/ sqrt5

S=3sqrt2*sqrt10*2/sqrt5= 12

6 0

2 years ago