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2 years ago

Given: WX congruent yz Prove: WYXZ

Mathematics

1 answer:

scZoUnD [109]2 years ago

3 0

Answer:

Segment Addition

Step-by-step explanation:

1. WX ≅ YZ Given

2. XY = XY Reflexive

3. WX + XY ≅ YZ + YZ Addition

4. WY ≅ XZ Substitution/ Segment addition

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3 years ago

Solve for u<br> 6≤ -9/4u

masya89 [10]

if you mean.

$6 \leqslant \frac{4u}{ - 9}$

then the answer would be,

$u \leqslant - \frac{27}{2}$

7 0

2 years ago

Read 2 more answers

A sporting goods store has 33 sets of golf clubs in stock. They get a shipment with 52 more sets of golf clubs. How many sets of

belka [17]

<h3>The golf club has 85 stes of golf clubs now</h3>

<em><u>Solution:</u></em>

Given that,

A sporting goods store has 33 sets of golf clubs in stock

They get a shipment with 52 more sets of golf clubs

To find: Number of golf clubs in the store now

From given,

Stock = 33 sets

shipment = 52 sets

Therefore,

Number of golf clubs in the store now = 33 + 52

Number of golf clubs in the store now = 85

Thus the golf club has 85 stes of golf clubs now

3 0

3 years ago

Find the quadratic equation for y=4x^2-5

marta [7]

Answer:

y=4/x^3

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

3 0

3 years ago