Find the value of X (in this picture)​

. ( m →(N v Q)) →((M→N) v (M→Q))

2.((T v U) → V) → (T → (U → V)) wut tyhe hell is this

Answer:

Greg is 28.

Step-by-step explanation:

Greg's age is y.

y - 5 = 23

y - 5 + 5 = 23 + 5

y = 28

Answer:

y = 2/5x + 1/5

Step-by-step explanation:

First, plug in the slope.

y = mx+b

y = 2/5x + b

Then, plug in the point given, in order to find b.

-1 = 2/5(-3) + b

-1 = -6/5 + b

1/5 = b

So the final equation is:

y = 2/5x + 1/5

Answer:

A.

Null:

H0: u= 5.65

Alternative:

H1: u > 5.65

B.

The test statistic

Xbar = 6.24

u = 5.65

S = 1.24

N = 57

T = 6.24-5.65/(1.24/√57)

= 0.59/(1.24/7.5498)

= 0.59/0.1642

Test statistic = 3.593

C.

With alpha = 0.05

Df = n-1 = 57-1 = 56

T critical = 1.673

D.

If t test > 1.673 reject the null hypothesis

3.593>1.673, reject the null hypothesis

We conclude that there is enough evidence that the technique helped the participants to learn more.

Answer:

"fair" srry im only in 8th so im d.u.m

Step-by-step explanation:

Answer:

13122

Step-by-step explanation:

a(n) = 6 * 3^(n -1)

a(8) = 6* 3^7

3^7 = 2187

a(8) =6 * 2187

a(8) = 13122

Answer:

B. 56° and 38°

Step-by-step explanation:

The sum of each of the two angles sums up to 90° ;

A.) 13° and 77°

Sum of A ; (13 + 77)° = 90°

B.) 56° + 38°

Sum = 56° + 38° = 94°

C) 37° and 53°

Sum = 37° + 53° = 90°

D.) 42° and 48°

Sum = 42° + 48° = 90°

Hence, the pair of degree measure which does not belong is B. 56° and 38°

Answer:

50/50

Step-by-step explanation:

It's random

Answer:

Option c (Upper tailed) is the correct choice.

Step-by-step explanation:

Given that:

The average attendance is:

= 74,900

We will have to test:

⇒ [tex]H_0:\mu \leq \mu_0[/tex]

or,

   [tex]H_0: \mu \leq 74,900[/tex]

Verses,

⇒ [tex]H_1: \mu> \mu_0[/tex]

or,

    [tex]H_1: \mu >74,900[/tex]

The other given alternatives aren't connected to the given scenario. So the above is the correct one.

Given:

ABC is an isosceles triangle in which AC =BC.

D and E are points on BC and AC such that CE=CD.

To prove:

Triangle ACD and BCE are congruent​.

Solution:

In triangle ACD and BCE,

[tex]AC=BC[/tex]                  (Given)

[tex]AC\cong BC[/tex]

[tex]\angle C\cong m\angle C[/tex]                  (Common angle)

[tex]CD=CE[/tex]                  (Given)

[tex]CD\cong CE[/tex]

In triangles ACD and BCE two corresponding sides and one included angle are congruent. So, the triangles are congruent by SAS congruence postulate.

[tex]\Delta ACD\cong \Delta BCE[/tex]           (SAS congruence postulate)

Hence proved.

Answer:

Given ABC is an isosceles triangle with AB=AC .D and E are the point on BC such that BE=CD

∴∠ABD=∠ACE (opposite angle of sides of a triangle ) ....(1)

Then BE−DE=CD−DE

ORBC=CE......................................(2)

In ΔABD and ΔACE

∠ABD=∠ACE ( From 1)

BC=CE (from 2)

AB=AC ( GIven)

∴ΔABD≅ΔACE

So AD=AE [henceproved]

Answer:

[tex]Shaded = 212.5cm^2[/tex]

Step-by-step explanation:

Given

See attachment

Required

The shaded region

First, calculate the area of the complete rectangle

[tex]Area =Length * Width[/tex]

[tex]Area = 14cm * 25cm[/tex]

[tex]Area = 350cm^2[/tex]

Next, calculate the area of the triangle

[tex]Area = \frac{1}{2} * Base * Height[/tex]

[tex]Area = \frac{1}{2} *25cm * 11cm[/tex]

[tex]Area = 137.5cm^2[/tex]

Subtract the area to calculate the shaded region area

[tex]Shaded = 350cm^2 - 137.5cm^2[/tex]

[tex]Shaded = 212.5cm^2[/tex]

Answer:

parallel

Step-by-step explanation:

the slope of each line is 5/6 which means the lines are parallel

Find the area of the top, which is a circle.

Area of a circle = pi x r^2

Area = 3.14 x 1.5^2 = 7.065

Now for the volume multiply the area of the top by the height:

7.065 x 12 = 84.78 cubic inches

Round to the nearest tenth: 84.8 cubic inches

Answer:

Volume of cylinder is 84.78 inches.

Step-by-step explanation:

Volume of cylinder = π × r ² × h

Where,

And we have given that

Substitute the values

Volume of cylinder = 3.14 × (1.5 inches) × 12 inches

= 3.14 × 2.25 inches² × 12 inches

multiply the values , we get

Volume of cylinder = 84.78 inches³.

84.8 inches³ is the round nearest to tenth.

Answer:

The point of intersection is:

[tex]\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)[/tex]

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

[tex]y=0.5x+5[/tex]

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Where (h, k) is the center and r is the radius.

Since our center is (0, 5), h = 0 and k = 5. The radius is 3. So, r = 3. Substitute:

[tex](x-0)^2+(y-5)^2=(3)^2[/tex]

Simplify:

[tex]x^2+(y-5)^2=9[/tex]

At the point where the two equations intersect, its x-coordinate and y-coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for x. So:

[tex]x^2+((0.5x+5)-5)^2=9[/tex]

Simplify:

[tex]x^2+(0.5x)^2=9[/tex]

Square:

[tex]x^2+0.25x^2=9[/tex]

Combine like terms:

[tex]\displaystyle 1.25x^2=\frac{5}{4}x^2=9[/tex]

Solve for x:

[tex]\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}[/tex]

Note that since we are looking for the point of intersection in QI, x should be positive. So, we can ignore the negative answer.

To find the y-coordinate, substitute the x-value back into either equation. Using the linear equation:

[tex]\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34[/tex]

So, the point of intersection in QI is:

[tex]\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)[/tex]

Answer:

1.043

Step-by-step explanation:

Decimal multiplier to increase by x% is computed as:-

100% + x%

The decimal multiplier to increase by 4.3% would be

100% +4.3% = 104.3%

To remove% we divide the number by 100, so the number will become

Hence, The decimal multiplier to increase by 4.3% = 1.043

Please Mark as Brainliest

Hope this Helps

Answer:

nnn bjbjjjjjnn

Step-by-step explanation:

Answer:

c. 108

Step-by-step explanation:

do 12×12 and get 144 then do 6×6 to get 36 then do 144-36 and you get 108. hope this helps mate.

Answer:

10 square inches

Step-by-step explanation:

that is the procedure above

Answer:

Area=10.167=10.2

Step-by-step explanation:

Area=1/2ab sin c

area= 1/2(7)(9.4)sin18

area = 10.167=10.2

this is for delta math

Answer:

4.605170185988091

Step-by-step explanation:

ln 100 = 4.605170185988091

e^4.605170185988091 = 100

Step-by-step explanation:

100=1 log 10(1)=0

I hope this is helpful for you

Answer:

Exact Form:

x = 2 √ 2 − 1 3 , − 2 √ 2 − 1 3

Decimal Form:

x = 2.49509379 , − 3.16176045

Step-by-step explanation:

Simplify the decimals as needed

Answer:

3408

Step-by-step explanation:

Answer:

ifeywifheb

Step-by-step explanation:

hkefewib

Answer:

x = 25

Step-by-step explanation:

Given a line parallel to a side of the triangle and intersecting the other 2 sides then it divides those sides proportionally, that is

[tex]\frac{40}{24}[/tex] = [tex]\frac{x}{15}[/tex] ( cross- multiply )

24x = 600 ( divide both sides by 24 )

x = 25

Answer:

Step-by-step explanation:

(0,2). y=mx+b or 2=-2 × 0+b, or solving for b: b=2-(-2)(0).

Answer:

the required equqtion is x-y=2

Answer:

The volume increases at 35.71cm^3/min

Step-by-step explanation:

Given

[tex]PV^{1.4} = C[/tex]

[tex]V = 400cm^3[/tex]

[tex]P =80kPa[/tex]

[tex]\frac{dP}{dt} =-10kPa/min[/tex]

Required

Rate at which volume increases

[tex]PV^{1.4} = C[/tex]       [tex]V = 400cm^3[/tex]       [tex]P =80kPa[/tex]

Differentiate: [tex]PV^{1.4} = C[/tex]  

[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = \frac{d}{dt}C[/tex]

By differentiating C, we have:

[tex]P*\frac{dV^{1.4}}{dt} +V^{1.4}*\frac{dP}{dt} = 0[/tex]

Rewrite as:

[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} + V^{1.4}*\frac{dP}{dt} = 0[/tex]

Solve for [tex]\frac{dV}{dt}[/tex]

[tex]P*(1.4)*V^{0.4}* \frac{dV}{dt} =- V^{1.4}*\frac{dP}{dt}[/tex]

[tex]\frac{dV}{dt} =- \frac{V^{1.4}*\frac{dP}{dt} }{P*(1.4)*V^{0.4}}[/tex]

Substitute values

[tex]\frac{dV}{dt} =- \frac{400^{1.4}*-10 }{80*(1.4)*400^{0.4}}[/tex]

[tex]\frac{dV}{dt} =\frac{400*10 }{80*1.4}[/tex]

[tex]\frac{dV}{dt} =\frac{4000 }{112}[/tex]

[tex]\frac{dV}{dt} =35.71cm^3/min[/tex]

9514 1404 393

Answer:

  (b)  (x – 4)^2 + (y – 6)^2 = 16

Step-by-step explanation:

The equation for a circle with center (h, k) and radius r is ...

  (x -h)^2 +(y -k)^2 = r^2

You want the equation for a circle with center (4, 6) and radius 4. That equation is ...

  (x -4)^2 +(y -6)^2 = 16

Answer:

B; (x – 4)2 + (y – 6)2 = 16

Step-by-step explanation:

correct on edge