What is: 8-3x=128 Please solve this :)

X+2y=6 and x-y=3 solve by linear equations by substitution

nevsk [136]

X+2y=6
x-y=3x=6-2y
x=3+y

6-2y=3+y
3=3y
y=1

x=3+1
x=4
Answer: (4;1)

3 0

3 years ago

A monthly contribution of $200 is deposited into an interest-bearing account.If the account has a 4.8% interest rate and compoun

bearhunter [10]

Answer:

22848 is your answer after you subtract the interest rate

Step-by-step explanation:

8 0

2 years ago

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

1 year ago

Please assist me with this problem

n200080 [17]

Answer:

The answer to your question is 90 dB

Step-by-step explanation:

Data

I = 10⁻³

I⁰ = 10⁻¹²

Formula

Loudness = 10log ( $\frac{I}{Io}$ )

Process

1.- To solve this problem, just substitute the values in the equation and do the operations.

2.- Substitution

Loudness = 10 log $(\frac{10^{-3}}{10^{-12}} )$

3.- Simplify

Loudness = 10log (1 x 10⁹)

Loudness = 10(9)

Loudness = 90

7 0

3 years ago

Substitute the given values into the formula, and solve for the unknown variable.

Alla [95]

Answer:

$h=\frac{1}{6}$

Step-by-step explanation:

Given:

$A=\frac{1}{2h} (B+b)\\$

$A=21, B=2, b=5 ,h= ?$

We have to find the value of 'h'

So, putting the given values in the equation to find the value of 'h'.

$A=\frac{1}{2h} (B+b)$

Implementing the given values in the given equations:

$21=\frac{1}{2h}(2+5)\\\\ 21=\frac{1}{2h} (7)\\\\\frac{1}{2h}=\frac{21}{7}\\\\ \frac{1}{2h}=3 \\\\h=\frac{1}{6}$

So, the value of 'h' is

$h=\frac{1}{6}$

3 0

3 years ago