All categories

4 years ago

In this quadrilateral x= A) 56. B) 67 C) 78 D) 89

Mathematics

1 answer:

Oxana [17]4 years ago

6 0

the correct answer is c 78

Step-by-step explanation:

lit can't be a because to small and can't be b because also to small. and and 92 is uptus and this is an acute angle

You might be interested in

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

3 years ago

Exact solutions. X^2-5x-1=0

Romashka [77]

Answer:

x= (5±√29)÷2

Step-by-step explanation:

Quadratic formula:

x= (-b±√b²-4ac)÷2a

7 0

3 years ago

Aleena expands 2xx+3-5(x-2) as 2x2+3-5x-10. Has she expanded it correctly? Please identify the mistakes and write the correct wa

Nimfa-mama [501]

The given expression is :

2xx+3-5(x-2)

We know that, $x{\cdot}x=x^2$

$=2x^2+3-5(x-2)\\\\=2x^2+3+(-5)(x)+(-5)(-2)\\\\=2x^2+3-5x+10$

Aleena expands 2xx+3-5(x-2) as 2x²+3-5x-10. She is mistaken because the sign before 10 should be +10 instead of -10. Hence, the correct expanded form is 2x²+3-5x+10. Hence, this is the required solution.

5 0

3 years ago

The table below shows the ages of employees over 50 at a company and the amount of money they have in their retirement accounts.

sp2606 [1]

Answer:

Step-by-step explanation:

plot your graph and then you will see there is no line which shows there is a correlation. they're too scattered

4 0

4 years ago

Hey guys can u tell me if the top is correct if its not can u show me step by step what to do and what the answer will be and ca

ASHA 777 [7]

You have the right idea and you are close to the correct answer. However that's not what your teacher is looking for in terms of steps.

The starting inequality is $b + 8 \ge 20$
She starts off selling 8 buckets. Then she sells b more to get a total of b+8
This total must be 20 or larger which is why I set b+8 greater than or equal to 20

Solve for b by subtracting 8 from both sides

$b + 8 \ge 20$

$b + 8-8 \ge 20-8$

$b \ge 12$

So she needs to sell at least 12 more buckets to reach her goal

If you graphed this on a number line, then you would draw a closed circle at 12. Then shade to the right of the close circle.

6 0

3 years ago