find the size of each of the unknown angles
plz give me solution​

Answer:

x=1

Step-by-step explanation:

log_4(x + 3) + log_4x = 1

We know that loga(b) + loga(c) = loga(bc)

log_4(x + 3)x = 1

Raise each side to the base of 4

4^log_4(x + 3)x = 4^1

(x+3)x = 4

x^2 +3x = 4

Subtract 4 from each side

x^2 +3x -4 = 0

Factor

(x+4) (x-1) =0

Using the zero product property

x= -4  x=1

But x cannot be negative since logs cannot be negative

x=1

Answer:

A..  x = 1.

Step-by-step explanation:

log_4(x + 3) + log_4x = 1

log_4 x(x + 3) = log_4 4

Removing the logs:

x(x + 3) = 4

x^2 + 3x - 4 = 0

(x + 4)(x - 1) = 0

x = 1, -4.

We can ignore the -4 as there is no log of a negative.

Answer:

11.3

Step-by-step explanation:

a = AC × sin(A)/sin(B)

Now <A =180-25-48 = 107

a = 5×sin(107)/sin(25)

a ≈ 11.3

Answered by GAUTHMATH

Answer:

The additive inverse of -61 is 61. For additive inverse just reverse the sign.

Step-by-step explanation:

here ,

length = 8m

height=4m

Rate of plastering 4 walls = 15/m^2

Cost of plastering 4 wall = Rs 1560

Area of floor = Cost of plastering / Rate of plastering

= 1560/15

= 104

again,

Area of four wall = 2h(l+b)

or, 104= 2×4(8+b)

or, 104 =8(8+b)

or, 104/8=8+ b

or,13 - 8= b

or, b= 5

again,

Area of floor= l×b

= 8×5

= 40

again,

Rate of carpeting the floor= 300/ m^2

Area of floor = 40 m^2

cost of carpeting=Rate × Area

= 300×40

= 12000

Answer:

5/3 x 5/3 x 5/3

125/27

5^3/3^3

3 x (5/3)

Step-by-step explanation:

(5/3)^3 = 5×5×5/3×3×3 = 125/27.

Answer:

-26 sqrt(3)

Step-by-step explanation:

-12 sqrt(12) - 2 sqrt(3)

Rewriting

-12 sqrt(4*3) - 2 sqrt(3)

We know sqrt(ab) = sqrt(a)sqrt(b)

-12 sqrt(4)sqrt(3) - 2 sqrt(3)

-12 (2) sqrt(3) - 2 sqrt(3)

-24 sqrt(3) - 2 sqrt(3)

-26 sqrt(3)

[tex] \large \tt{{❃ \: S \: O \: L \: U \: T \: I \: O \: N : }}[/tex]

[tex] \large{ \tt{❁ \:x + x + 98 = 180 \degree \: [ Sum\: of \: angle \: of \: a \: triangle ]}}[/tex]

[tex] \large{ \tt{⟶2x + 98 \degree= 180 \degree}}[/tex]

[tex] \large{ \tt{⟶ \: 2x = 180 \degree - 98 \degree}}[/tex]

[tex] \large{ \tt{⟶ \: 2x = 82 \degree}}[/tex]

[tex] \large{ \tt{ ⟶x = \frac{82 \degree}{2} }}[/tex]

[tex] \large{ \tt{⟶ \: x = 41 \degree}}[/tex]

[tex] \large{ \tt{ ↔\angle \: 1 = x \degree = \boxed{41 \degree}}}[/tex] [ Being alternate angles ]

- Alternate angles are the non-adjacent interiors pair of angles lying to the opposite side of a transversal when it intersects two straight line segments. Alternate angles form ' Z ' shape.

Answer:

Step-by-step explanation:

Answer:

i think you just extend the coordinates to the side, except the right point, by 3, and then the bottom ones go down by 3, and the top one goes up by 3

Step-by-step explanation:

Answer:

x = 55°

Step-by-step explanation:

The sum of the angles in a triangle is 180°.

x + 75 + 50 = 180

x + 125 = 180

x = 55

Therefore, x = 55°.

Answer:

x = 55

Step-by-step explanation:

the angles of a triangle add up to 180 degrees. So far we have 125 taken up.

We do 180 - 125 = 55.

x = 55

Step-by-step explanation:

[tex] \frac{dy}{dx} = \frac{1}{ {(6 - 3x)}^{4} } \times {4(6 - 3x)}^{3} ( - 3) \\ = \frac{ - 12 {(6 - 3x)}^{3} }{ {(6 - 3x)}^{4} } \\ = - \frac{12}{6 - 3x} [/tex]

I hope I'm correct. I've never learnt differentiation for log and exponents before

In the diagram, ∠5 is adjacent to ∠EHD.

Angle

Angles are formed when two rays intersect at a point. An angle is also formed when to lines intersect each other, thereby the two lines share a common endpoint.

Adjacent angles are two angles that have a common side and a common vertex (corner point). They are placed side by side to each other.

In the diagram, ∠5 is adjacent to ∠EHD.

Find out more on Angle at: https://brainly.com/question/25770607

Question 8

a^2 + 7a + 12

= (a+3)(a+4)

When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).

An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.

a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)

= (-7 +or- sqrt(1))/2

= -3 or -4

These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).

Question 10

-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18

Rewrite each fraction with a common denominator so you can combine the fractions into one.

= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18

= (-6(x - y) + 9(4x - 7y) - (x + y)) /18

Expand the brackets and collect like terms.

= (-6x + 6y + 36x - 63y - x - y)/18

= (29x - 58y)/18

= 29/18 x - 29/9 y

Answer:

In 26 seconds to complete a transaction, 2 products are being purchased.

Step-by-step explanation:

1 item  = 9 seconds

Time taken in all to check out = 8 seconds

Time taken to shop = 26 seconds

Now as check out process takes 8 seconds, so the

Time left to ACTUALLY SHOP =  Total Time  - Time Used to check out

                                                  = 26 seconds = 8 seconds = 18 seconds

Shopping of 1 item = 8 seconds

Shopping of 2 items  = 2 x ( Time taken in 1 item) = 2 x  9 = 18 seconds

So, in 18 seconds, 2 clothing item can be selected.

Hence,in 26 seconds to complete a transaction, 2 products are being purchased.

they don't share any points

that's one thing, but I don't know what your options are of course.

see screenshot for illustration of the inequalities

Part 1

[tex]\left(\frac{g}{h}\right)(x) = \frac{g(x)}{h(x)}\\\\\left(\frac{g}{h}\right)(x) = \frac{3x-5}{-2x^2+7}\\\\\left(\frac{g}{h}\right)(3) = \frac{3(3)-5}{-2(3)^2+7}\\\\\left(\frac{g}{h}\right)(3) = \frac{4}{-11}\\\\\left(\frac{g}{h}\right)(3) = -\frac{4}{11}\\\\[/tex]

Answer:  -4/11

====================================================

Part 2

Set the denominator function equal to zero and solve for x to find which values to kick out of the domain.

[tex]h(x) = 0\\\\-2x^2+7 = 0\\\\7 = 2x^2\\\\2x^2 = 7\\\\x^2 = 7/2\\\\x^2 = 3.5\\\\x = \sqrt{3.5} \ \text{ or } x = -\sqrt{3.5}\\\\[/tex]

This shows that if x is equal to either of those values, then the denominator h(x) will be zero. These are the values to kick out of the domain to prevent a division by zero error. Any other value of x is valid in the domain.

Answer: [tex]x = \sqrt{3.5} \text{ and } x = -\sqrt{3.5}\\\\[/tex]

Answer:

30 =x

Step-by-step explanation:

The angles are correcting angle and corresponding angles are equal when the lines are parallel

55 = x+25

Subtract 25 from each side

55-25 = x+25

30 =x

First go to the y intercept (or the b in y=mx+b) look for the slope and plot the points on the graph they're talking about e.g. #23 the y-intercept is 6 go to the 6 on the y axis, and then look at the slope (x), so it goes up and over to the right since it's positive by 1

Answer:

Let be a rational complex number of the form [tex]z = \frac{a + i\,b}{c + i\,d}[/tex], we proceed to show the procedure of resolution by algebraic means:

1) [tex]\frac{a + i\,b}{c + i\,d}[/tex]   Given.

2) [tex]\frac{a + i\,b}{c + i\,d} \cdot 1[/tex] Modulative property.

3) [tex]\left(\frac{a+i\,b}{c + i\,d} \right)\cdot \left(\frac{c-i\,d}{c-i\,d} \right)[/tex]   Existence of additive inverse/Definition of division.

4) [tex]\frac{(a+i\,b)\cdot (c - i\,d)}{(c+i\,d)\cdot (c - i\,d)}[/tex]   [tex]\frac{x}{y}\cdot \frac{w}{z} = \frac{x\cdot w}{y\cdot z}[/tex]  

5) [tex]\frac{a\cdot (c-i\,d) + (i\,b)\cdot (c-i\,d)}{c\cdot (c-i\,d)+(i\,d)\cdot (c-i\,d)}[/tex]  Distributive and commutative properties.

6) [tex]\frac{a\cdot c + a\cdot (-i\,d) + (i\,b)\cdot c +(i\,b) \cdot (-i\,d)}{c^{2}-c\cdot (i\,d)+(i\,d)\cdot c+(i\,d)\cdot (-i\,d)}[/tex] Distributive property.

7) [tex]\frac{a\cdot c +i\,(-a\cdot d) + i\,(b\cdot c) +(-i^{2})\cdot (b\cdot d)}{c^{2}+i\,(c\cdot d)+[-i\,(c\cdot d)] +(-i^{2})\cdot d^{2}}[/tex] Definition of power/Associative and commutative properties/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction.

8) [tex]\frac{(a\cdot c + b\cdot d) +i\cdot (b\cdot c -a\cdot d)}{c^{2}+d^{2}}[/tex] Definition of imaginary number/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction/Distributive, commutative, modulative and associative properties/Existence of additive inverse/Result.

Step-by-step explanation:

Let be a rational complex number of the form [tex]z = \frac{a + i\,b}{c + i\,d}[/tex], we proceed to show the procedure of resolution by algebraic means:

1) [tex]\frac{a + i\,b}{c + i\,d}[/tex]   Given.

2) [tex]\frac{a + i\,b}{c + i\,d} \cdot 1[/tex] Modulative property.

3) [tex]\left(\frac{a+i\,b}{c + i\,d} \right)\cdot \left(\frac{c-i\,d}{c-i\,d} \right)[/tex]   Existence of additive inverse/Definition of division.

4) [tex]\frac{(a+i\,b)\cdot (c - i\,d)}{(c+i\,d)\cdot (c - i\,d)}[/tex]   [tex]\frac{x}{y}\cdot \frac{w}{z} = \frac{x\cdot w}{y\cdot z}[/tex]  

5) [tex]\frac{a\cdot (c-i\,d) + (i\,b)\cdot (c-i\,d)}{c\cdot (c-i\,d)+(i\,d)\cdot (c-i\,d)}[/tex]  Distributive and commutative properties.

6) [tex]\frac{a\cdot c + a\cdot (-i\,d) + (i\,b)\cdot c +(i\,b) \cdot (-i\,d)}{c^{2}-c\cdot (i\,d)+(i\,d)\cdot c+(i\,d)\cdot (-i\,d)}[/tex] Distributive property.

7) [tex]\frac{a\cdot c +i\,(-a\cdot d) + i\,(b\cdot c) +(-i^{2})\cdot (b\cdot d)}{c^{2}+i\,(c\cdot d)+[-i\,(c\cdot d)] +(-i^{2})\cdot d^{2}}[/tex] Definition of power/Associative and commutative properties/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction.

8) [tex]\frac{(a\cdot c + b\cdot d) +i\cdot (b\cdot c -a\cdot d)}{c^{2}+d^{2}}[/tex] Definition of imaginary number/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction/Distributive, commutative, modulative and associative properties/Existence of additive inverse/Result.

The net cost after discounts will be $ 2,450.

Given that Economy Hardware Store ordered items retailing for $ 2,500, and they received a chain discount of 5/20/2, the following calculation must be performed to find the net cost, knowing that the net cost is equal to the initial cost minus the discounts made about it:

First, the discount percentage must be calculated.

5/20/2 = X

4/2 = X

2 = X

Then, this percentage must be subtracted from the initial value.

2500 x (1-0.02) = X

2500 x 0.98 = X

2450 = X

Therefore, the net cost after discounts will be $ 2,450.

Learn more in https://brainly.com/question/17003148.

Answer:

12

Step-by-step explanation:

Add 10l to 12l to13l to11l to 14l=60l the divide 60l by the number of houses which will be 12 and there is your correct answer

Answer: He invested 8,00,000.

Step-by-step explanation:

R = 5%

A = 840,000

T = 1 year

so

so

I = A-P

so

I = PTR/100

or, A-P = (p*1*5)/100

or, 840,000-P = 5p/100

or, 8,40,00,000-100P = 5p

or, 8,40,00,000 = 105P

so, p = 8,40,8,40,00,000

so, P = 8,00,000

Answer:

-12x-9+50-5=0

-12x+41-5=0

-12x+36=0

-12x=0-36

x= -36/-12

x = 3 Answer...

hope it helps

Answer:

[tex]x=3[/tex]

Step-by-step explanation:

[tex]3(-4x - 3) + 50 - 5= 0[/tex]

⇒ Subtract 50- 5 from both sides:-

[tex]3\left(-4x-3\right)+50-5-\left(50-5\right)=0-\left(50-5\right)[/tex]

[tex]3\left(-4x-3\right)=-45[/tex]

⇒ Divide both sides by 3:-

[tex]\frac{3\left(-4x-3\right)}{3}=\frac{-45}{3}[/tex]

[tex]-4x-3=-15[/tex]

⇒ Add 3 to both sides:-

[tex]-4x-3+3=-15+3[/tex]

[tex]-4x=-12[/tex]

⇒ Divide both sides by -4:-

[tex]\frac{-4x}{-4}=\frac{-12}{-4}[/tex]

[tex]x=3[/tex]

OAmalOHopeO

Answer:

[tex]\angle 1 + \angle 2 = 180^o[/tex]

Step-by-step explanation:

Given

[tex]\angle 1[/tex] and [tex]\angle 2[/tex]

Required

The relationship between them [tex]\angle 1[/tex] and [tex]\angle 2[/tex]

From the question, we understand that [tex]\angle 1[/tex] and [tex]\angle 2[/tex] are supplementary

Supplementary angles add up to 180.

So, the relationship between [tex]\angle 1[/tex] and [tex]\angle 2[/tex] is:

[tex]\angle 1 + \angle 2 = 180^o[/tex]

Answer:

Volume: 52 Units Squared

Surface Area: 94 units.

Step-by-step explanation:

The volume is relatively simple to find. Just subtract the original volume by the 2x2x2 cube's volume. The original volume is 60. The cube's volume is 2x2x2 which is 8. 60-8=52.

The surface area is harder to find. Try to envision the corner of the rectangle being cut out. We see that each side of the cube has a surface area of 2x2 which is 4. In the picture, we see that three sides of the rectangle has been partially removed. But since each side of the cube has an equal surface area, it is safe to minus 3 of the sides that has been partially removed by 3. However, since that it is the corner, the "dent" that the cube made in the rectangle also needed to be counted. As we said, each of the sides of a cube has a surface area of 4, so since that the dent has 3 sides, we see that the surface area of the dent is 4x3 which is 12. Now we need to count the unaffected sides of the rectangle. There are three of them. Just multiply the edges to find the surface area of each side. Add all of the values up: 11+16+12+8+15+12+20=94 units.

Answer:

m BC = 100

Step-by-step explanation:

Since the angle is at the center, the arc has the same measurement as the angle

m BC = 100

This triangle has base 16 therefore the sides must be 17cm and 17 cm

When we make a altitude it divides it into two right triangles and there is a property in which the altitude of the isoceles triangle divides the base in 2 equal halves

So the side of the right triangle will be x , 8 , 17

Using pythgoreus theorem

x²+8²=17²

x = √225

x = 15

So the altitude is 15 cm

Must click thanks and mark brainliest

Answer:

30.80

Step-by-step explanation:

We can use a ratio to solve

4 tickets              7 tickets

-------------------  = ----------------

17.60 dollars           x dollars

Using cross products

4x = 17.60 * 7

4x =123.2

Divide each side by 4

4x/4 = 123.2/4

x=30.8

Answer:

4, 8, 16, 32, 64

Step-by-step explanation:

The nth term of a geometric sequence is

[tex]a_{n}[/tex] = a₁[tex](r)^{n-1}[/tex]

Given

a₇ = 256 and a₁₀ = 2048 , then

a₁ [tex]r^{6}[/tex] = 256 → (1)

a₁ [tex]r^{9}[/tex] = 2048 → (2)

Divide (2) by (1)

[tex]\frac{a_{1}r^{9} }{a_{1}r^{6} }[/tex] = [tex]\frac{2048}{256}[/tex]

r³ = 8 ( take the cube root of both sides )

r = [tex]\sqrt[3]{8}[/tex] = 2

Substitute r = 2 into (1)

a₁ × [tex]2^{6}[/tex] = 256

a₁ × 64 = 256 ( divide both sides by 64 )

a₁ = 4

Then

a₁ = 4

a₂ = 2a₁ = 2 × 4 = 8

a₃ = 2a₂ = 2 × 8 = 16

a₄ = 2a₃ = 2 × 16 = 32

a₅ = 2a₄ = 2 × 32 = 64